Electrostatics MCQs with Answers – Part 7 (NEET Physics)

Explanation: \(\textbf{Determine the two capacitor fields:}\) The capacitors have equal free plate-charge magnitudes and equal areas. The vacuum field in P is \[ E_P=\frac{Q}{\varepsilon_0A} \] Capacitor S contains dielectric constant \(K=3\): \[ E_S=\frac{Q}{3\varepsilon_0A} \] Therefore, \[ E_P=3E_S \] \(\textbf{Use the dipole work relation:}\) The potential energy of a dipole in a uniform field is \[ U=-pE\cos\theta \] Work done by the electrostatic field during rotation from \(\theta_i\) to \(\theta_f\) is \[ W=pE(\cos\theta_f-\cos\theta_i) \] \(\textbf{Calculate work in capacitor P:}\) The rotation is from \(60^\circ\) to \(0^\circ\): \[ W_P=pE_P(\cos0^\circ-\cos60^\circ) \] \[ W_P=pE_P\left(1-\frac12\right) \] \[ W_P=\frac12pE_P \] \(\textbf{Calculate work in capacitor S:}\) The rotation is from \(120^\circ\) to \(60^\circ\): \[ W_S=pE_S(\cos60^\circ-\cos120^\circ) \] \[ W_S=pE_S\left(\frac12+\frac12\right) \] \[ W_S=pE_S \] \(\textbf{Compare the rotational works:}\) \[ \frac{W_P}{W_S}=\frac{pE_P/2}{pE_S} \] \[ \frac{W_P}{W_S}=\frac{E_P}{2E_S} \] Using \(E_P=3E_S\), \[ \frac{W_P}{W_S}=\frac32 \] \(\textbf{Compare the initial torques:}\) The torque magnitude is \[ \tau=pE|\sin\theta| \] For P, \[ \tau_{P,i}=pE_P\sin60^\circ \] \[ \tau_{P,i}=\frac{\sqrt3}{2}pE_P \] For S, \[ \tau_{S,i}=pE_S\sin120^\circ \] \[ \tau_{S,i}=\frac{\sqrt3}{2}pE_S \] Thus, \[ \frac{\tau_{P,i}}{\tau_{S,i}}=\frac{E_P}{E_S} \] \[ \frac{\tau_{P,i}}{\tau_{S,i}}=3 \] \(\textbf{Physical interpretation:}\) The dielectric reduces the field in S because both capacitors have the same fixed free charge. The two prescribed dipole rotations cover different changes in cosine, so their work ratio is not simply the field ratio. Their initial sine magnitudes, however, are equal, making the initial torque ratio equal to the field ratio. \(\textbf{Final answer:}\) \[ \boxed{\frac{W_P}{W_S}=\frac32,\qquad \frac{\tau_{P,i}}{\tau_{S,i}}=3} \]

Explanation: \(\textbf{Identify the conductor's effect on capacitance:}\) The electric field inside the conducting slab is zero at electrostatic equilibrium. Therefore, the potential difference occurs only across the remaining vacuum gaps. Their combined thickness is \[ x-a \] \(\textbf{Write capacitance as a function of separation:}\) \[ C(x)=\frac{\varepsilon_0A}{x-a} \] Therefore, \[ \frac1{C(x)}=\frac{x-a}{\varepsilon_0A} \] Expanding, \[ \frac1C=\frac{x}{\varepsilon_0A}-\frac{a}{\varepsilon_0A} \] \(\textbf{Interpret the experimental graph slope:}\) Comparing with the measured straight line, \[ \frac1C=sx-as \] the slope is \[ s=\frac1{\varepsilon_0A} \] Therefore, \[ A=\frac1{\varepsilon_0s} \] \(\textbf{Determine the initial reciprocal capacitance:}\) At separation \(x_i=2a\), \[ \frac1{C_i}=s(2a)-as \] \[ \frac1{C_i}=as \] \(\textbf{Determine the final reciprocal capacitance:}\) At separation \(x_f=3a\), \[ \frac1{C_f}=s(3a)-as \] \[ \frac1{C_f}=2as \] \(\textbf{Express energy using reciprocal capacitance:}\) The capacitor is isolated, so \(Q\) remains fixed. Its stored energy is \[ U=\frac{Q^2}{2C} \] Therefore, \[ U=\frac{Q^2}{2}\left(\frac1C\right) \] \(\textbf{Calculate the initial and final energies:}\) Initially, \[ U_i=\frac{Q^2}{2}(as) \] \[ U_i=\frac{Q^2as}{2} \] Finally, \[ U_f=\frac{Q^2}{2}(2as) \] \[ U_f=Q^2as \] \(\textbf{Determine the energy increase:}\) \[ \Delta U=U_f-U_i \] \[ \Delta U=Q^2as-\frac{Q^2as}{2} \] \[ \Delta U=\frac{Q^2as}{2} \] \(\textbf{Determine external mechanical work:}\) There is no connected battery, and the process is quasistatic and lossless. Therefore, \[ W_{\mathrm{ext}}=\Delta U \] \[ W_{\mathrm{ext}}=\frac{Q^2as}{2} \] \(\textbf{Physical interpretation:}\) The reciprocal-capacitance graph reveals both the effective geometry and the conductor's excluded field thickness. At fixed charge, stored energy is proportional to reciprocal capacitance. The positive external work is needed to separate attracting plates while increasing the field-filled vacuum volume. \(\textbf{Final answer:}\) \[ \boxed{A=\frac1{\varepsilon_0s},\qquad \Delta U=\frac{Q^2as}{2},\qquad W_{\mathrm{ext}}=\frac{Q^2as}{2}} \]

Explanation: \(\textbf{Identify the two contributions to dipole potential energy:}\) The two real charges interact with each other through Coulomb's law. They also interact with the externally maintained uniform electric field. These are different energy contributions and must be distinguished. \(\textbf{Evaluate mutual Coulomb energy:}\) The separation between the real dipole charges is \(2a\). Their mutual potential energy is \[ U_{\mathrm{mutual}}=\frac{k(+q)(-q)}{2a} \] \[ U_{\mathrm{mutual}}=-\frac{kq^2}{2a} \] The rotation does not change their separation. Therefore, \[ \Delta U_{\mathrm{mutual}}=0 \] \(\textbf{Determine the capacitor's electric field:}\) For isolated large vacuum parallel plates, \[ E=\frac{Q}{\varepsilon_0A} \] This field is treated as fixed during the dipole rotation. \(\textbf{Determine the real dipole moment magnitude:}\) A real dipole formed by charges \(\pm q\) separated by \(2a\) has moment \[ p=q(2a) \] \[ p=2qa \] \(\textbf{Evaluate the initial external-field interaction energy:}\) Initially, the dipole moment is parallel to the field: \[ \theta_i=0^\circ \] Thus, \[ U_{\mathrm{ext},i}=-pE\cos0^\circ \] \[ U_{\mathrm{ext},i}=-pE \] \(\textbf{Evaluate the final interaction energy:}\) After rotation through \(180^\circ\), \[ \theta_f=180^\circ \] Therefore, \[ U_{\mathrm{ext},f}=-pE\cos180^\circ \] \[ U_{\mathrm{ext},f}=+pE \] \(\textbf{Calculate external rotational work:}\) The change in total dipole potential energy is \[ \Delta U=\Delta U_{\mathrm{mutual}}+\Delta U_{\mathrm{ext}} \] \[ \Delta U=0+[pE-(-pE)] \] \[ \Delta U=2pE \] For quasistatic rotation, \[ W_{\mathrm{ext}}=\Delta U \] Substituting \(p=2qa\), \[ W_{\mathrm{ext}}=4qaE \] \[ W_{\mathrm{ext}}=\frac{4qaQ}{\varepsilon_0A} \] \(\textbf{Determine maximum torque:}\) The external-field torque magnitude is \[ \tau=pE|\sin\theta| \] It becomes maximum at \(90^\circ\): \[ \tau_{\max}=pE \] \[ \tau_{\max}=2qa\frac{Q}{\varepsilon_0A} \] \[ \tau_{\max}=\frac{2qaQ}{\varepsilon_0A} \] \(\textbf{Physical interpretation:}\) The charges' mutual attraction does not change its potential energy because their separation is fixed. The rotational work arises from changing the dipole's orientation relative to the external capacitor field. The electrostatic torque vanishes at the parallel and antiparallel orientations, even though the corresponding interaction energies differ. \(\textbf{Final answer:}\) \[ \boxed{\Delta U_{\mathrm{mutual}}=0,\qquad W_{\mathrm{ext}}=\frac{4qaQ}{\varepsilon_0A},\qquad \tau_{\max}=\frac{2qaQ}{\varepsilon_0A}} \]

Explanation: Procedure P is physically consistent because an isolated capacitor retains its free plate charge. Doubling its capacitance therefore halves its voltage through \(V=Q/C\). At fixed charge, stored energy is inversely proportional to capacitance, so its energy also halves. Procedure Q is consistent because the ideal battery maintains the initial voltage. When capacitance doubles at fixed voltage, the charge increases by the same factor. The stored energy also doubles because \(U=\frac12CV^2\). Procedure R imposes a different but physically possible condition: the source adjusts charge so that stored electrostatic energy remains fixed. Using \(U=Q^2/(2C)\), doubling capacitance at unchanged energy requires the charge to increase by the factor \(\sqrt2\). The voltage then becomes \(V_0/\sqrt2\), consistent with the claimed final values. Procedure S is inconsistent because it simultaneously claims fixed charge, unchanged voltage and doubled capacitance. These three claims cannot satisfy \(Q=CV\) for a nonzero charge and voltage. In addition, an isolated capacitor cannot acquire additional free plate charge without a charge-transfer path. The comparison shows that fixed charge, fixed voltage and fixed stored energy are distinct experimental constraints. Each can be legitimate, but each produces a different relationship between final charge and voltage when capacitance changes. The consistency check must be applied to the complete set of physical quantities rather than considering one stated value in isolation. Procedures P, Q and R are consistent, whereas S is not.

Explanation: \(\textbf{Determine the initial series capacitance:}\) The two equal capacitors are initially in series: \[ C_{\mathrm{eq},i}=\frac{C\cdot C}{C+C} \] \[ C_{\mathrm{eq},i}=\frac C2 \] \(\textbf{Calculate the initial battery charge:}\) \[ Q_i=C_{\mathrm{eq},i}V_0 \] \[ Q_i=\frac{CV_0}{2} \] \(\textbf{Determine the original junction voltage:}\) Both series capacitors have the same capacitance and charge magnitude. They therefore have equal voltage drops: \[ V_{AX}=V_{XB}=\frac{V_0}{2} \] Since \(V_B=0\), \[ V_{X,i}=\frac{V_0}{2} \] \(\textbf{Interpret the closed switch:}\) Closing the switch connects X directly to B: \[ V_{X,f}=V_B=0 \] The second capacitor now has zero potential difference: \[ V_{XB,f}=0 \] It therefore stores no charge or electrostatic energy in its final equilibrium state. \(\textbf{Determine the final charge on the first capacitor:}\) The first capacitor remains between A at \(V_0\) and X at zero potential. Thus, \[ Q_f=C(V_A-V_X) \] \[ Q_f=CV_0 \] \(\textbf{Calculate additional charge supplied by the positive battery terminal:}\) Initially, the A-side plate carries \(CV_0/2\). Finally, it carries \(CV_0\): \[ \Delta Q=CV_0-\frac{CV_0}{2} \] \[ \Delta Q=\frac{CV_0}{2} \] \(\textbf{Determine the original stored energy:}\) \[ U_i=\frac12C_{\mathrm{eq},i}V_0^2 \] \[ U_i=\frac12\left(\frac C2\right)V_0^2 \] \[ U_i=\frac{CV_0^2}{4} \] \(\textbf{Determine the final stored energy:}\) Only the first capacitor stores energy: \[ U_f=\frac12CV_0^2 \] \(\textbf{Calculate work done by the battery:}\) The battery maintains \(V_0\) while supplying charge \(\Delta Q\): \[ W_{\mathrm{battery}}=V_0\Delta Q \] \[ W_{\mathrm{battery}}=\frac{CV_0^2}{2} \] \(\textbf{Determine the energy increase:}\) \[ \Delta U=U_f-U_i \] \[ \Delta U=\frac{CV_0^2}{2}-\frac{CV_0^2}{4} \] \[ \Delta U=\frac{CV_0^2}{4} \] \(\textbf{Calculate the switching loss:}\) The battery work is divided between increased stored electrostatic energy and dissipation: \[ W_{\mathrm{battery}}=\Delta U+H \] Therefore, \[ H=\frac{CV_0^2}{2}-\frac{CV_0^2}{4} \] \[ H=\frac{CV_0^2}{4} \] \(\textbf{Physical interpretation:}\) Grounding the intermediate junction changes the circuit's equilibrium conditions. The original equal-series-charge relationship no longer describes the final topology. The first capacitor becomes directly connected across the battery, while the second is short-circuited. Although the total stored energy increases, some energy is dissipated as the switch closes. \(\textbf{Final answer:}\) \[ \boxed{\Delta Q=\frac{CV_0}{2},\qquad U_f=\frac12CV_0^2,\qquad H=\frac14CV_0^2} \]

Explanation: \(\textbf{Express the potential on the charge axis:}\) For a point at coordinate \(x\) distinct from either charge, \[ V(x)=\frac{4kQ}{|x|}-\frac{kQ}{|x-3a|} \] The absolute values are essential because electric potential depends on positive distances, while charge signs are retained algebraically. \(\textbf{Find the zero-potential point between the charges:}\) For \[ 0\lt x\lt3a \] the distances from the positive and negative charges are \(x\) and \(3a-x\). The zero-potential condition is \[ \frac{4kQ}{x}-\frac{kQ}{3a-x}=0 \] Cancelling \(kQ\), \[ \frac4x=\frac1{3a-x} \] \[ 4(3a-x)=x \] \[ 12a=5x \] Therefore, \[ x_A=\frac{12a}{5} \] This lies within the stipulated interval. \(\textbf{Find the zero-potential point beyond the negative charge:}\) For \(x\gt3a\), the distances are \(x\) and \(x-3a\). Thus, \[ \frac4x=\frac1{x-3a} \] \[ 4(x-3a)=x \] \[ 3x=12a \] Therefore, \[ x_B=4a \] This lies to the right of the negative charge. \(\textbf{Determine the field direction at A:}\) Between the charges, the field of \(+4Q\) points away from it, along \(+\hat i\). The field of \(-Q\) points towards that charge, also along \(+\hat i\). Therefore, the resultant field at A points along \(+\hat i\). \(\textbf{Determine the field direction at B:}\) At \(x=4a\), the field of \(+4Q\) points along \(+\hat i\), with magnitude \[ E_+=\frac{4kQ}{(4a)^2}=\frac{kQ}{4a^2} \] The field of \(-Q\) points along \(-\hat i\), with magnitude \[ E_-=\frac{kQ}{a^2} \] Since \(E_-\gt E_+\), the resultant field points along \(-\hat i\). \(\textbf{Determine electrostatic work:}\) Both A and B have zero electric potential: \[ V_A=V_B=0 \] For a positive test charge \(q\), \[ W_{A\to B}=q(V_A-V_B) \] \[ W_{A\to B}=0 \] \(\textbf{Physical interpretation:}\) Two points can have equal electric potentials while the electric fields at those points have different directions. Electric potential is a scalar sum, whereas electric field is a vector sum. The zero work follows from equal endpoint potentials, not from a vanishing field along the path. \(\textbf{Final answer:}\) \[ \boxed{x_A=\frac{12a}{5},\qquad x_B=4a,\qquad \vec E_A\parallel+\hat i,\qquad \vec E_B\parallel-\hat i,\qquad W_{A\to B}=0} \]

Explanation: \(\textbf{Interpret the first graph segment:}\) From \(x=0\) to \(x=a\), the potential increases linearly from zero to \(3V_0\). Therefore, \[ \frac{\Delta V}{\Delta x}=\frac{3V_0}{a} \] The electric field is the negative potential gradient: \[ E_x=-\frac{\Delta V}{\Delta x} \] Thus, \[ E_{0\lt x\lt a}=-\frac{3V_0}{a} \] The field points along negative \(x\). \(\textbf{Determine the positive particle's energy change:}\) A positive charge moving along increasing \(x\) in this region moves against the electric force. Its kinetic energy decreases as the electric potential rises. At a coordinate \(x\) in the first region, \[ V(x)=\frac{3V_0x}{a} \] The increase in potential energy is \[ \Delta U=qV(x)=\frac{3qV_0x}{a} \] \(\textbf{Find the first turning point:}\) The particle starts with kinetic energy \[ K_i=2qV_0 \] At its first turning point, kinetic energy is zero. Energy conservation gives \[ 2qV_0=\frac{3qV_0x_{\mathrm{turn}}}{a} \] Cancelling \(qV_0\), \[ 2=\frac{3x_{\mathrm{turn}}}{a} \] Therefore, \[ x_{\mathrm{turn}}=\frac{2a}{3} \] Because this coordinate lies before \(x=a\), the particle turns without entering the second region. \(\textbf{Determine the field in the second region:}\) Between \(x=a\) and \(x=3a\), the potential falls from \(3V_0\) to \(V_0\). Its slope is \[ \frac{\Delta V}{\Delta x}=\frac{V_0-3V_0}{3a-a} \] \[ \frac{\Delta V}{\Delta x}=-\frac{V_0}{a} \] Therefore, \[ E_{x\gt a}=+\frac{V_0}{a} \] The field points along \(+\hat i\). \(\textbf{Calculate work on the negative test charge:}\) The separate charge moves from \(x=3a\), where \(V=V_0\), to \(x=0\), where \(V=0\). Electrostatic work is \[ W=q_t(V_i-V_f) \] Using \(q_t=-q\), \[ W=(-q)(V_0-0) \] \[ W=-qV_0 \] \(\textbf{Physical interpretation:}\) The graph slope determines the field direction, while potential differences determine work. The launched positive particle cannot cross a potential barrier whose required energy exceeds its initial kinetic energy. The negative test charge experiences negative field work during the specified movement because its potential energy increases. \(\textbf{Final answer:}\) \[ \boxed{x_{\mathrm{turn}}=\frac{2a}{3},\qquad \vec E_{x\gt a}\parallel+\hat i,\qquad W=-qV_0} \]

Explanation: \(\textbf{Use the square's common source distance:}\) Each vertex lies at distance \[ r=\sqrt{a^2+a^2} \] \[ r=\sqrt2a \] from the origin. \(\textbf{Determine the initial potential:}\) Electric potential is a scalar. Initially, the four source charges have algebraic sum \[ Q+Q-Q-Q=0 \] Because every source has the same distance from the origin, \[ V_i=\frac{k}{\sqrt2a}(Q+Q-Q-Q) \] \[ V_i=0 \] \(\textbf{Determine the initial field direction:}\) The fields of the two upper positive charges point away from those charges, towards the lower half-plane. The fields of the lower negative charges point towards those charges, also towards the lower half-plane. Horizontal field components cancel by left-right symmetry. Thus the initial field points along \(-\hat j\). \(\textbf{Calculate the initial field magnitude:}\) The field magnitude from one vertex is \[ E_1=\frac{kQ}{r^2} \] \[ E_1=\frac{kQ}{2a^2} \] Its vertical component magnitude is \[ E_{1y}=E_1\cos45^\circ \] \[ E_{1y}=\frac{kQ}{2\sqrt2a^2} \] All four vertical components point downward: \[ \vec E_i=-4E_{1y}\hat j \] \[ \vec E_i=-\frac{\sqrt2kQ}{a^2}\hat j \] \(\textbf{Treat the changed source as an additional charge:}\) Changing the charge at \((a,a)\) from \(+Q\) to \(-Q\) is equivalent, for superposition purposes, to adding charge \(-2Q\) at that vertex. \(\textbf{Calculate the potential change:}\) \[ \Delta V=\frac{k(-2Q)}{\sqrt2a} \] \[ \Delta V=-\frac{\sqrt2kQ}{a} \] Since \(V_i=0\), \[ V_f=-\frac{\sqrt2kQ}{a} \] \(\textbf{Calculate the additional field vector:}\) The field of the equivalent \(-2Q\) charge points from the origin towards \((a,a)\). Its magnitude is \[ |\Delta\vec E|=\frac{k(2Q)}{2a^2} \] \[ |\Delta\vec E|=\frac{kQ}{a^2} \] The equal components are \[ \Delta\vec E=\frac{kQ}{\sqrt2a^2}(\hat i+\hat j) \] \(\textbf{Determine the final field:}\) \[ \vec E_f=\vec E_i+\Delta\vec E \] \[ \vec E_f=-\frac{2kQ}{\sqrt2a^2}\hat j+\frac{kQ}{\sqrt2a^2}(\hat i+\hat j) \] Therefore, \[ \vec E_f=\frac{kQ}{\sqrt2a^2}(\hat i-\hat j) \] \(\textbf{Calculate work in bringing the test charge:}\) The initial test-charge potential at infinity is zero. The final electric potential at the origin is \(V_f\). Thus, \[ W_{\mathrm{field}}=q(0-V_f) \] \[ W_{\mathrm{field}}=+\frac{\sqrt2kQq}{a} \] \(\textbf{Physical interpretation:}\) The initial square has zero potential at its centre but a nonzero electric field. Changing one source charge breaks the left-right field symmetry and produces a horizontal field component. The scalar potential change and vector field change must be determined separately. \(\textbf{Final answer:}\) \[ \boxed{V_f=-\frac{\sqrt2kQ}{a},\qquad \vec E_f=\frac{kQ}{\sqrt2a^2}(\hat i-\hat j),\qquad W_{\mathrm{field}}=\frac{\sqrt2kQq}{a}} \]

Explanation: \(\textbf{Use the potential measured at the origin:}\) Both source charges lie at distance \(a\) from the origin. Therefore, \[ V(0)=\frac{kq_1}{a}+\frac{kq_2}{a} \] The measured value is \(kQ/a\): \[ \frac{k(q_1+q_2)}a=\frac{kQ}a \] Hence, \[ q_1+q_2=Q \] \(\textbf{Use the second potential measurement:}\) At \(x=2a\), the distances from \(q_1\) and \(q_2\) are \(3a\) and \(a\), respectively. Thus, \[ V(2a)=\frac{kq_1}{3a}+\frac{kq_2}{a} \] The measurement gives \[ \frac{kq_1}{3a}+\frac{kq_2}{a}=\frac{kQ}{2a} \] Cancelling \(k/a\), \[ \frac{q_1}{3}+q_2=\frac Q2 \] \(\textbf{Solve the source charges:}\) From the first equation, \[ q_2=Q-q_1 \] Substituting, \[ \frac{q_1}{3}+Q-q_1=\frac Q2 \] \[ -\frac{2q_1}{3}=-\frac Q2 \] Therefore, \[ q_1=\frac{3Q}{4} \] and \[ q_2=\frac Q4 \] \(\textbf{Determine the electric field at the origin:}\) Both sources are positive. The field from \(q_1\), situated at \(x=-a\), points along \(+\hat i\). The field from \(q_2\), situated at \(x=+a\), points along \(-\hat i\). Therefore, \[ \vec E(0)=\frac{kq_1}{a^2}\hat i-\frac{kq_2}{a^2}\hat i \] \[ \vec E(0)=\frac{k(q_1-q_2)}{a^2}\hat i \] Substituting, \[ \vec E(0)=\frac{k(3Q/4-Q/4)}{a^2}\hat i \] \[ \vec E(0)=\frac{kQ}{2a^2}\hat i \] \(\textbf{Determine work on the negative test charge:}\) The endpoint potentials are \[ V_i=\frac{kQ}{a} \] \[ V_f=\frac{kQ}{2a} \] Electrostatic work is \[ W_{\mathrm{field}}=q_t(V_i-V_f) \] Using \(q_t=-q\), \[ W_{\mathrm{field}}=(-q)\left(\frac{kQ}{a}-\frac{kQ}{2a}\right) \] \[ W_{\mathrm{field}}=-\frac{kQq}{2a} \] \(\textbf{Physical interpretation:}\) Two scalar potential measurements determine the unknown source charges. Once their magnitudes are known, field direction requires vector reasoning rather than scalar addition. The negative test charge loses kinetic energy during the specified movement if no other force performs work. \(\textbf{Final answer:}\) \[ \boxed{q_1=\frac{3Q}{4},\qquad q_2=\frac Q4,\qquad \vec E(0)=\frac{kQ}{2a^2}\hat i,\qquad W_{\mathrm{field}}=-\frac{kQq}{2a}} \]

Explanation: \(\textbf{Identify axial dipole-potential dependence:}\) For an ideal electric dipole, the potential at distance \(r\) on its positive axial side is \[ V(r)=\frac{kp}{r^2} \] Thus, axial potential decreases as the inverse square of distance. \(\textbf{Relate the dipole moment to the measurement:}\) At A, \[ V_A=\frac{kp}{a^2} \] The given value is \(V_0\): \[ \frac{kp}{a^2}=V_0 \] Therefore, \[ kp=V_0a^2 \] \(\textbf{Calculate the potential at B:}\) Point B lies at distance \(2a\): \[ V_B=\frac{kp}{(2a)^2} \] \[ V_B=\frac{kp}{4a^2} \] Substituting the measured relation, \[ V_B=\frac{V_0}{4} \] \(\textbf{Identify axial electric-field dependence:}\) The magnitude of the field of an ideal dipole on its axis is \[ E(r)=\frac{2kp}{r^3} \] Therefore, axial field magnitude decreases as the inverse cube of distance. \(\textbf{Compare field strengths:}\) At A, \[ E_A=\frac{2kp}{a^3} \] At B, \[ E_B=\frac{2kp}{(2a)^3} \] \[ E_B=\frac{2kp}{8a^3} \] Hence, \[ \frac{E_A}{E_B}=8 \] \(\textbf{Calculate electrostatic work:}\) The transported charge is negative: \[ q_t=-q \] Its work from A to B is \[ W_{\mathrm{field}}=q_t(V_A-V_B) \] Substituting, \[ W_{\mathrm{field}}=(-q)\left(V_0-\frac{V_0}{4}\right) \] \[ W_{\mathrm{field}}=-\frac{3qV_0}{4} \] \(\textbf{Physical interpretation:}\) Both potential and field decrease with distance from a dipole, but their radial power laws differ. The negative test charge moves from higher to lower potential, increasing its potential energy. Consequently, the dipole field performs negative work on it during the outward movement. \(\textbf{Final answer:}\) \[ \boxed{V_B=\frac{V_0}{4},\qquad \frac{E_A}{E_B}=8,\qquad W_{\mathrm{field}}=-\frac{3qV_0}{4}} \]

Explanation: \(\textbf{Interpret the interior-potential measurement:}\) The electric field inside an empty charged conducting spherical shell is zero. However, its electric potential is not generally zero. The entire interior and conducting surface share the same potential. Thus, \[ V_0=\frac{kQ}{R} \] \(\textbf{Use the exterior-field measurement:}\) Immediately outside the charged spherical shell, \[ E_0=\frac{kQ}{R^2} \] \(\textbf{Eliminate the unknown charge:}\) Divide the potential relation by the field relation: \[ \frac{V_0}{E_0}=\frac{kQ/R}{kQ/R^2} \] Simplifying, \[ \frac{V_0}{E_0}=R \] Therefore, \[ R=\frac{V_0}{E_0} \] \(\textbf{Determine the total shell charge:}\) From the potential relation, \[ Q=\frac{V_0R}{k} \] Substituting the inferred radius, \[ Q=\frac{V_0^2}{kE_0} \] Using \[ k=\frac1{4\pi\varepsilon_0} \] we obtain \[ Q=\frac{4\pi\varepsilon_0V_0^2}{E_0} \] \(\textbf{Determine the endpoint potentials:}\) At the shell surface, \[ V_i=V_0 \] At radial distance \(2R\), \[ V_f=\frac{kQ}{2R} \] Therefore, \[ V_f=\frac{V_0}{2} \] \(\textbf{Calculate field work on the negative charge:}\) \[ W_{\mathrm{field}}=q_t(V_i-V_f) \] With \(q_t=-q\), \[ W_{\mathrm{field}}=(-q)\left(V_0-\frac{V_0}{2}\right) \] \[ W_{\mathrm{field}}=-\frac{qV_0}{2} \] \(\textbf{Check the physical direction:}\) The shell is positively charged, so its exterior field points radially outward. A negative test charge experiences an inward force. Moving it outward therefore requires positive external work and involves negative field work. \(\textbf{Physical interpretation:}\) Interior potential and surface field provide different information about the same charged shell. Their measured ratio determines the shell radius directly. The constant nonzero interior potential is compatible with the zero electric field inside the shell. \(\textbf{Final answer:}\) \[ \boxed{R=\frac{V_0}{E_0},\qquad Q=\frac{4\pi\varepsilon_0V_0^2}{E_0},\qquad W_{\mathrm{field}}=-\frac{qV_0}{2}} \]

Explanation: \(\textbf{Use potential superposition for spherical shells:}\) A uniformly charged thin spherical shell produces constant potential throughout its interior. For a shell of radius \(r_s\) and charge \(Q_s\), the interior potential is \[ V_{\mathrm{inside}}=\frac{kQ_s}{r_s} \] At an exterior point, its potential is \[ V_{\mathrm{outside}}=\frac{kQ_s}{r} \] These contributions add algebraically. \(\textbf{Calculate the potential at A:}\) Point A lies inside both spherical shells. The inner shell contributes \[ V_{A,1}=\frac{k(-2Q)}R \] \[ V_{A,1}=-\frac{2kQ}{R} \] The outer shell contributes \[ V_{A,2}=\frac{kQ}{2R} \] Therefore, \[ V_A=-\frac{2kQ}{R}+\frac{kQ}{2R} \] \[ V_A=-\frac{3kQ}{2R} \] \(\textbf{Locate B relative to each shell:}\) Point B has radial coordinate \[ r_B=\frac{3R}{2} \] It is outside the inner shell but inside the outer shell. \(\textbf{Calculate the potential at B:}\) The inner shell contributes \[ V_{B,1}=\frac{k(-2Q)}{3R/2} \] \[ V_{B,1}=-\frac{4kQ}{3R} \] The outer shell contributes the same constant interior potential: \[ V_{B,2}=\frac{kQ}{2R} \] Thus, \[ V_B=-\frac{4kQ}{3R}+\frac{kQ}{2R} \] \[ V_B=-\frac{5kQ}{6R} \] \(\textbf{Determine the field direction at B:}\) A spherical Gaussian surface of radius \(3R/2\) encloses the inner shell's charge \(-2Q\). The outer shell produces zero field within its spherical interior. Therefore, the resultant field at B points radially inward towards the negatively charged inner shell. Its magnitude is \[ E_B=\frac{k(2Q)}{(3R/2)^2} \] \[ E_B=\frac{8kQ}{9R^2} \] \(\textbf{Calculate field work on the positive test charge:}\) \[ W_{\mathrm{field}}=q(V_A-V_B) \] Substituting, \[ W_{\mathrm{field}}=q\left(-\frac{3kQ}{2R}+\frac{5kQ}{6R}\right) \] \[ W_{\mathrm{field}}=-\frac{2kQq}{3R} \] \(\textbf{Physical interpretation:}\) The centre and the inner-shell surface share the same potential because the field inside the inner shell is zero. Between the shells, the electric field is nonzero and directed inward. Moving a positive test charge outward through this region increases its potential energy, so the field performs negative work. \(\textbf{Final answer:}\) \[ \boxed{V_A=-\frac{3kQ}{2R},\qquad V_B=-\frac{5kQ}{6R},\qquad E_B=\frac{8kQ}{9R^2}\ \text{inward},\qquad W_{\mathrm{field}}=-\frac{2kQq}{3R}} \]

Explanation: \(\textbf{Apply the angular dipole-potential relation:}\) At distance \(r\) and polar angle \(\theta\), the potential of an ideal dipole is \[ V(r,\theta)=\frac{kp\cos\theta}{r^2} \] The distance is the same for A, B and C. Their potentials therefore differ only through the cosine factor. \(\textbf{Calculate the potential at A:}\) \[ V_A=\frac{kp\cos60^\circ}{r^2} \] \[ V_A=\frac{kp}{2r^2} \] \(\textbf{Calculate the potential at B:}\) \[ V_B=\frac{kp\cos90^\circ}{r^2} \] \[ V_B=0 \] \(\textbf{Calculate the potential at C:}\) \[ V_C=\frac{kp\cos120^\circ}{r^2} \] \[ V_C=-\frac{kp}{2r^2} \] \(\textbf{Determine external work during A to B:}\) For quasistatic transport of a positive charge, \[ W_{\mathrm{ext}}=\Delta U=q(V_f-V_i) \] Therefore, \[ W_{\mathrm{ext},AB}=q(V_B-V_A) \] \[ W_{\mathrm{ext},AB}=-\frac{kpq}{2r^2} \] \(\textbf{Determine external work during B to C:}\) \[ W_{\mathrm{ext},BC}=q(V_C-V_B) \] \[ W_{\mathrm{ext},BC}=-\frac{kpq}{2r^2} \] Both external works are equal and negative. Hence, their ratio is \[ W_{\mathrm{ext},AB}:W_{\mathrm{ext},BC}=1:1 \] \(\textbf{Determine net field work from A to C:}\) The total potential-energy change is \[ \Delta U_{AC}=q(V_C-V_A) \] \[ \Delta U_{AC}=q\left(-\frac{kp}{2r^2}-\frac{kp}{2r^2}\right) \] \[ \Delta U_{AC}=-\frac{kpq}{r^2} \] Therefore, \[ W_{\mathrm{field},AC}=-\Delta U_{AC} \] \[ W_{\mathrm{field},AC}=+\frac{kpq}{r^2} \] \(\textbf{Check the requested sign:}\) The two stages both move the positive test charge towards lower potential. The electric field therefore performs positive work over the complete A-to-C transport. The positive field-work sign is essential because electrostatic work equals the negative of the potential-energy change. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{ext},AB}:W_{\mathrm{ext},BC}=1:1,\qquad W_{\mathrm{field},AC}=+\frac{kpq}{r^2}} \]

Explanation: \(\textbf{Determine the scalar potential at O:}\) All three vertices lie at the same distance \(R\) from O. Potential contributions add algebraically: \[ V_O=\frac{k(+Q)}R+\frac{k(+Q)}R+\frac{k(-Q)}R \] Therefore, \[ V_O=\frac{kQ}{R} \] The measured potential is \(V_0\): \[ V_0=\frac{kQ}{R} \] \(\textbf{Use triangular symmetry for the electric field:}\) Let position vectors from O to the vertices be \(\vec r_1,\vec r_2,\vec r_3\). Their magnitudes are all \(R\), and their vector sum is zero: \[ \vec r_1+\vec r_2+\vec r_3=0 \] The third vertex carries \(-Q\). \(\textbf{Write the electric-field superposition:}\) The field at O from a charge \(q_i\) at \(\vec r_i\) is \[ \vec E_i=-\frac{kq_i}{R^3}\vec r_i \] Therefore, \[ \vec E_O=-\frac{kQ}{R^3}(\vec r_1+\vec r_2-\vec r_3) \] \(\textbf{Apply the vertex-vector relation:}\) Since \[ \vec r_1+\vec r_2=-\vec r_3 \] we obtain \[ \vec E_O=-\frac{kQ}{R^3}(-2\vec r_3) \] \[ \vec E_O=\frac{2kQ}{R^3}\vec r_3 \] Thus, the field points from O towards the negative-charge vertex. \(\textbf{Calculate field magnitude:}\) Because \(|\vec r_3|=R\), \[ E_0=\frac{2kQ}{R^2} \] \(\textbf{Eliminate the unknown source charge:}\) From the potential measurement, \[ kQ=V_0R \] Substitute into the field expression: \[ E_0=\frac{2V_0R}{R^2} \] \[ E_0=\frac{2V_0}{R} \] Therefore, \[ R=\frac{2V_0}{E_0} \] \(\textbf{Determine the charge magnitude:}\) Using \[ Q=\frac{V_0R}{k} \] we obtain \[ Q=\frac{V_0}{k}\frac{2V_0}{E_0} \] \[ Q=\frac{2V_0^2}{kE_0} \] \(\textbf{Physical interpretation:}\) The net potential depends on the algebraic charge sum because the source distances are equal. The electric field depends on the angular positions and vector directions of all three contributions. Measurements of both potential and field therefore identify the geometric scale and source-charge magnitude. \(\textbf{Final answer:}\) \[ \boxed{R=\frac{2V_0}{E_0},\qquad Q=\frac{2V_0^2}{kE_0},\qquad \vec E_O\ \text{points towards the negative-charge vertex}} \]

Explanation: \(\textbf{Calculate the potential at O:}\) The origin lies at distance \(a\) from each positive charge. Therefore, \[ V_O=\frac{kQ}{a}+\frac{kQ}{a} \] \[ V_O=\frac{2kQ}{a} \] \(\textbf{Express the potential on the perpendicular bisector:}\) Let P have coordinate \((0,y)\), with \(y\gt0\). Its distance from either charge is \[ r=\sqrt{a^2+y^2} \] Hence, \[ V_P=\frac{2kQ}{\sqrt{a^2+y^2}} \] \(\textbf{Apply the measured potential ratio:}\) The measurement gives \[ V_P=\frac{V_O}{2} \] Therefore, \[ \frac{2kQ}{\sqrt{a^2+y^2}}=\frac{kQ}{a} \] Cancelling \(kQ\), \[ \sqrt{a^2+y^2}=2a \] Squaring, \[ a^2+y^2=4a^2 \] \[ y^2=3a^2 \] Since P lies on the positive \(y\)-axis, \[ y=\sqrt3a \] \(\textbf{Determine the field direction at P:}\) The horizontal components of the fields from the two positive charges cancel by symmetry. Their vertical components reinforce along \(+\hat j\). \(\textbf{Calculate the field magnitude:}\) At P, the distance from each source is \(2a\). The field magnitude from one charge is \[ E_1=\frac{kQ}{(2a)^2} \] \[ E_1=\frac{kQ}{4a^2} \] The vertical component is \[ E_{1y}=E_1\frac{y}{r} \] \[ E_{1y}=\frac{kQ}{4a^2}\frac{\sqrt3a}{2a} \] \[ E_{1y}=\frac{\sqrt3kQ}{8a^2} \] Both vertical components add: \[ \vec E_P=2E_{1y}\hat j \] \[ \vec E_P=\frac{\sqrt3kQ}{4a^2}\hat j \] \(\textbf{Calculate electrostatic work from O to P:}\) The potential at P is \[ V_P=\frac{kQ}{a} \] Thus, \[ W_{\mathrm{field}}=q(V_O-V_P) \] \[ W_{\mathrm{field}}=q\left(\frac{2kQ}{a}-\frac{kQ}{a}\right) \] \[ W_{\mathrm{field}}=\frac{kQq}{a} \] \(\textbf{Physical interpretation:}\) The origin has zero electric field by symmetry but a nonzero positive potential. At P, the electric field is nonzero and points away from the two charges. The positive test charge moves towards lower potential, so the field performs positive work. \(\textbf{Final answer:}\) \[ \boxed{P=(0,\sqrt3a),\qquad \vec E_P=\frac{\sqrt3kQ}{4a^2}\hat j,\qquad W_{\mathrm{field}}=\frac{kQq}{a}} \]

Explanation: \(\textbf{Interpret the horizontal graph section:}\) A uniformly charged spherical shell produces zero electric field throughout its empty interior. Consequently, its electric potential is constant for \[ r\lt R \] The graph gives this constant as \(V_0\). Thus, \[ V(R/2)=V_0 \] \(\textbf{Use the exterior potential relation:}\) Outside the shell, \[ V(r)=\frac{kQ}{r} \] At its surface, \[ V_0=\frac{kQ}{R} \] Therefore, \[ kQ=V_0R \] \(\textbf{Calculate the potential at }3R\textbf{:}\) \[ V(3R)=\frac{kQ}{3R} \] Substituting, \[ V(3R)=\frac{V_0R}{3R} \] \[ V(3R)=\frac{V_0}{3} \] \(\textbf{Determine electrostatic work:}\) The positive test charge moves from potential \(V_0\) to \(V_0/3\). The electrostatic work is \[ W_{\mathrm{field}}=q(V_i-V_f) \] \[ W_{\mathrm{field}}=q\left(V_0-\frac{V_0}{3}\right) \] \[ W_{\mathrm{field}}=\frac{2qV_0}{3} \] \(\textbf{Compare the one-sided electric fields at the surface:}\) Immediately inside the shell, \[ E(R^-)=0 \] Immediately outside, \[ E(R^+)=\frac{kQ}{R^2} \] Using \(kQ=V_0R\), \[ E(R^+)=\frac{V_0}{R} \] The exterior field is nonzero while the interior field is zero. The ratio \[ \frac{E(R^+)}{E(R^-)} \] therefore involves division by zero and is not a finite defined ratio. \(\textbf{Physical interpretation:}\) The electric potential remains continuous across an ideal charged spherical shell, even though the electric field has different one-sided values. The graph's constant interior potential does not imply zero potential energy for a test charge placed inside. The work depends only on the initial and final potentials. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{field}}=\frac{2qV_0}{3},\qquad E(R^-)=0,\qquad E(R^+)=\frac{V_0}{R},\qquad V(3R)=\frac{V_0}{3}} \]

Explanation: \(\textbf{Express axial ring potential:}\) Every charge element on the ring is at the same distance from an axial point. At coordinate \(x\), that distance is \[ r=\sqrt{R^2+x^2} \] Hence, the electric potential is \[ V(x)=\frac{kQ}{\sqrt{R^2+x^2}} \] \(\textbf{Express the axial electric field:}\) The transverse field components cancel by circular symmetry. The axial field magnitude for \(x\gt0\) is \[ E(x)=\frac{kQx}{(R^2+x^2)^{3/2}} \] \(\textbf{Use the measured field-to-potential ratio:}\) At \(x=a\), \[ \frac{E(a)}{V(a)}= \frac{kQa/(R^2+a^2)^{3/2}}{kQ/(R^2+a^2)^{1/2}} \] Cancelling common factors, \[ \frac{E(a)}{V(a)}=\frac{a}{R^2+a^2} \] The measurement gives \[ \frac{a}{R^2+a^2}=\frac1{3a} \] Cross-multiplying, \[ 3a^2=R^2+a^2 \] Therefore, \[ R^2=2a^2 \] \[ R=\sqrt2a \] \(\textbf{Determine the ring charge:}\) The measured potential at A is \[ V_0=\frac{kQ}{\sqrt{R^2+a^2}} \] Substituting \(R^2=2a^2\), \[ V_0=\frac{kQ}{\sqrt{3a^2}} \] \[ V_0=\frac{kQ}{\sqrt3a} \] Therefore, \[ Q=\frac{\sqrt3aV_0}{k} \] \(\textbf{Determine potential at the ring centre:}\) At \(x=0\), \[ V(0)=\frac{kQ}{R} \] Substituting, \[ V(0)=\frac{k(\sqrt3aV_0/k)}{\sqrt2a} \] \[ V(0)=\sqrt{\frac32}V_0 \] \(\textbf{Check the field behaviour:}\) At the centre, axial electric field is zero by symmetry. However, the ring potential there is positive and greater than its potential at the off-centre point A. This distinction is consistent with the inferred geometry. \(\textbf{Physical interpretation:}\) The potential and field of a uniformly charged ring have different axial dependence. Their ratio eliminates the unknown total charge and permits the ring radius to be inferred. The potential measurement then fixes the source charge, followed by the central potential. \(\textbf{Final answer:}\) \[ \boxed{R=\sqrt2a,\qquad Q=\frac{\sqrt3aV_0}{k},\qquad V(0)=\sqrt{\frac32}V_0} \]

Explanation: \(\textbf{Calculate the potential at the midpoint:}\) Point P lies at equal distance \(a\) from both source charges. Its potential is \[ V_P=\frac{kQ}{a}-\frac{kQ}{a} \] \[ V_P=0 \] \(\textbf{Determine the potential at S:}\) Point S lies at \(x=3a\). Its distances from the positive and negative charges are \(3a\) and \(a\), respectively. Therefore, \[ V_S=\frac{kQ}{3a}-\frac{kQ}{a} \] \[ V_S=-\frac{2kQ}{3a} \] \(\textbf{Determine the field of the positive source at S:}\) The field of \(+Q\) points away from the charge, along positive \(x\): \[ \vec E_+=+\frac{kQ}{(3a)^2}\hat i \] \[ \vec E_+=+\frac{kQ}{9a^2}\hat i \] \(\textbf{Determine the field of the negative source at S:}\) The field of \(-Q\) points towards it, along negative \(x\): \[ \vec E_-=-\frac{kQ}{a^2}\hat i \] \(\textbf{Add the field contributions:}\) \[ \vec E_S=\vec E_++\vec E_- \] \[ \vec E_S=\left(\frac{kQ}{9a^2}-\frac{kQ}{a^2}\right)\hat i \] \[ \vec E_S=-\frac{8kQ}{9a^2}\hat i \] \(\textbf{Calculate the electrostatic work:}\) For the positive test charge, \[ W_{\mathrm{field}}=q(V_P-V_S) \] Substituting, \[ W_{\mathrm{field}}=q\left(0+\frac{2kQ}{3a}\right) \] \[ W_{\mathrm{field}}=+\frac{2kQq}{3a} \] \(\textbf{Distinguish field and potential at P:}\) At the midpoint, the potentials due to \(+Q\) and \(-Q\) cancel. However, both electric-field contributions point towards the negative charge. Thus, the resultant field at P is nonzero despite its zero potential. \(\textbf{Physical interpretation:}\) The test charge moves between points of different electric potential, so the field performs nonzero work. The field direction at S follows vector superposition, while the work follows scalar potential difference. These calculations are independent and need not have the same sign convention. \(\textbf{Final answer:}\) \[ \boxed{V_P=0,\qquad \vec E_S=-\frac{8kQ}{9a^2}\hat i,\qquad W_{\mathrm{field}}=\frac{2kQq}{3a}} \]

Explanation: Statement I is correct because the electrostatic field is conservative. The work done on a test charge depends only on the potential difference between its endpoints. Every point on S has the same electric potential \(V_0\). Therefore, the potential difference between any two points on S is zero. The field work is zero even if the chosen path leaves the surface before reaching the second point, provided the electrostatic source configuration remains fixed. Statement II is incorrect because constant potential on a surface does not imply that the electric field vanishes there. A charged conducting sphere, for example, has constant potential over its surface while supporting a nonzero electric field immediately outside it. The field can be perpendicular to an equipotential surface without having a tangential component. Statement III is correct because Gauss's law relates the total flux through a closed surface to its net enclosed charge. The surface being equipotential does not force the enclosed net charge to vanish. A spherical equipotential surface surrounding a positive point charge provides a direct example with nonzero outward flux. Statement IV is correct because a nonzero tangential electric-field component would produce a potential change along the surface. That would contradict the stated equipotential condition. Thus the nonzero electrostatic field at a smooth equipotential surface is normal to that surface. The essential distinction is between zero potential difference along a surface and zero electric-field magnitude. Statements I, III and IV are correct.

Explanation: \(\textbf{Determine the uniform field:}\) The capacitor is isolated and vacuum-filled. Its free surface charge density is \[ \sigma=\frac QA \] The electric-field magnitude between the large plates is \[ E=\frac{\sigma}{\varepsilon_0} \] Therefore, \[ E=\frac{Q}{\varepsilon_0A} \] The field points along \(+\hat i\). \(\textbf{Relate potential to position:}\) In a uniform field along positive \(x\), \[ V(x_1)-V(x_2)=E(x_2-x_1) \] The electric potential decreases as \(x\) increases. The potential is independent of \(y\) within the ideal uniform-field region. \(\textbf{Compare P and R:}\) Both P and R have the same \(x\)-coordinate: \[ x_P=x_R=\frac d4 \] Therefore, \[ V_P=V_R \] \(\textbf{Calculate work during P to R:}\) The transported charge is \(-q\). Its electrostatic work is \[ W_{P\to R}=(-q)(V_P-V_R) \] Thus, \[ W_{P\to R}=0 \] \(\textbf{Determine the potential difference from R to S:}\) The two points have coordinates \[ x_R=\frac d4,\qquad x_S=\frac{3d}{4} \] Their displacement along the field is \[ \Delta x=x_S-x_R \] \[ \Delta x=\frac d2 \] Therefore, \[ V_R-V_S=E\frac d2 \] Substituting the capacitor field, \[ V_R-V_S=\frac{Qd}{2\varepsilon_0A} \] \(\textbf{Calculate electrostatic work during R to S:}\) \[ W_{R\to S}=(-q)(V_R-V_S) \] \[ W_{R\to S}=-\frac{qQd}{2\varepsilon_0A} \] \(\textbf{Determine total field work:}\) The two stages add: \[ W_{\mathrm{field,total}}=W_{P\to R}+W_{R\to S} \] \[ W_{\mathrm{field,total}}=-\frac{qQd}{2\varepsilon_0A} \] \(\textbf{Calculate total external quasistatic work:}\) The external agent transports the test charge without changing its kinetic energy. Therefore, \[ W_{\mathrm{ext}}=-W_{\mathrm{field,total}} \] \[ W_{\mathrm{ext}}=+\frac{qQd}{2\varepsilon_0A} \] \(\textbf{Determine the direct P-to-S potential difference:}\) Because P and R are equipotential, \[ V_P-V_S=V_R-V_S \] Hence, \[ V_P-V_S=\frac{Qd}{2\varepsilon_0A} \] \(\textbf{Physical interpretation:}\) The displacement from P to R is perpendicular to the uniform electric field and causes no potential change. The subsequent displacement from R to S is parallel to the field and produces a potential drop. A negative test charge moving towards lower potential increases its potential energy, requiring positive external work. \(\textbf{Final answer:}\) \[ \boxed{W_{P\to R}=0,\qquad W_{R\to S}=-\frac{qQd}{2\varepsilon_0A},\qquad W_{\mathrm{ext}}=+\frac{qQd}{2\varepsilon_0A},\qquad V_P-V_S=\frac{Qd}{2\varepsilon_0A}} \]