Electrostatics MCQs with Answers – Part 4 (NEET Physics)

Explanation: \(\textbf{Separate the central-charge flux from the uniform-field flux:}\) The point charge lies at the exact centre of the cube. By symmetry, it contributes the same outward flux through each of the six faces: \[ \phi_Q=\frac{Q}{6\varepsilon_0} \] Let the unknown uniform field be \[ \vec E_u=E_x\hat i+E_y\hat j+E_z\hat k \] The uniform-field contributions through opposite parallel faces are equal in magnitude and opposite in sign. \(\textbf{Use the pair of }x\textbf{-face measurements:}\) For the \(+\hat i\) face, \[ \frac{Q}{6\varepsilon_0}+E_xa^2=6\Phi_0 \] For the \(-\hat i\) face, \[ \frac{Q}{6\varepsilon_0}-E_xa^2=-2\Phi_0 \] Adding, \[ \frac{Q}{3\varepsilon_0}=4\Phi_0 \] Thus, \[ Q=12\varepsilon_0\Phi_0 \] The central charge contributes \[ \phi_Q=\frac{12\varepsilon_0\Phi_0}{6\varepsilon_0} \] \[ \phi_Q=2\Phi_0 \] through every cube face. \(\textbf{Find the uniform field's }x\textbf{-component:}\) Subtract the central-charge contribution from the \(+\hat i\) face reading: \[ E_xa^2=6\Phi_0-2\Phi_0 \] \[ E_xa^2=4\Phi_0 \] Therefore, \[ E_x=\frac{4\Phi_0}{a^2} \] \(\textbf{Recover the remaining field components:}\) For the \(+\hat j\) face, \[ 2\Phi_0+E_ya^2=\Phi_0 \] Therefore, \[ E_y=-\frac{\Phi_0}{a^2} \] For the \(+\hat k\) face, \[ 2\Phi_0+E_za^2=5\Phi_0 \] Hence, \[ E_z=\frac{3\Phi_0}{a^2} \] The reconstructed uniform field is \[ \vec E_u=\frac{\Phi_0}{a^2}(4\hat i-\hat j+3\hat k) \] \(\textbf{Calculate the flux through the new surface:}\) After removing the point charge, only this uniform field remains. The area vector of the new square is \[ \vec A=a^2\hat n \] \[ \vec A=\frac{a^2}{3}(2\hat i+2\hat j+\hat k) \] Therefore, \[ \Phi_{\mathrm{open}}=\vec E_u\cdot\vec A \] Substituting, \[ \Phi_{\mathrm{open}}=\frac{\Phi_0}{3}[(4)(2)+(-1)(2)+(3)(1)] \] \[ \Phi_{\mathrm{open}}=\frac{\Phi_0}{3}(8-2+3) \] \[ \Phi_{\mathrm{open}}=3\Phi_0 \] \(\textbf{Physical check:}\) The opposite \(x\)-face measurements distinguish the central charge from the background field. The remaining face measurements recover the other field components. The final open-surface flux requires the complete vector field, not merely its magnitude. \(\textbf{Final answer:}\) \[ \boxed{Q=12\varepsilon_0\Phi_0,\qquad \Phi_{\mathrm{open}}=3\Phi_0} \]

Explanation: \(\textbf{Determine the initial field at P:}\) Initially, the wire lies along the \(z\)-axis. Point P is at perpendicular distance \(R\) from the wire. The field magnitude due to an infinite uniformly charged wire is \[ E_i=\frac{\lambda}{2\pi\varepsilon_0R} \] Because the wire is positive, the electric field at P points towards negative \(x\). The force on the negative test charge therefore points towards positive \(x\). Its initial magnitude is \[ F_i=qE_i \] \[ F_i=\frac{q\lambda}{2\pi\varepsilon_0R} \] \(\textbf{Identify the distance after translating the wire:}\) The translated wire lies along the line \(x=R/2,\ y=0\). The test charge remains at \(x=-R,\ y=0\). Its new perpendicular distance from the wire is \[ r_f=\frac R2-(-R) \] \[ r_f=\frac{3R}{2} \] The wire still lies to the right of P. Therefore, the electric-field direction at P remains towards negative \(x\). \(\textbf{Calculate the modified test-charge force:}\) \[ E_f=\frac{\lambda}{2\pi\varepsilon_0(3R/2)} \] \[ E_f=\frac{2\lambda}{3(2\pi\varepsilon_0R)} \] The test charge is unchanged, so \[ \frac{F_f}{F_i}=\frac{E_f}{E_i} \] \[ \frac{F_f}{F_i}=\frac23 \] The force remains directed towards positive \(x\), but its magnitude decreases. \(\textbf{Compare the enclosed charges:}\) The Gaussian cylinder remains fixed. Initially, it encloses a wire segment of length \(L\). After the parallel translation, the wire still passes completely through both flat end faces and remains within the curved boundary. Thus, the enclosed wire length remains \(L\). The enclosed charge is consequently unchanged: \[ Q_{\mathrm{enc},i}=\lambda L \] \[ Q_{\mathrm{enc},f}=\lambda L \] \(\textbf{Apply Gauss's law to both arrangements:}\) \[ \Phi_i=\frac{\lambda L}{\varepsilon_0} \] \[ \Phi_f=\frac{\lambda L}{\varepsilon_0} \] Therefore, \[ \frac{\Phi_f}{\Phi_i}=1 \] \(\textbf{Distinguish flux from local field:}\) The off-axis wire destroys the original cylindrical symmetry about the Gaussian-cylinder axis. Its local field is no longer constant over the curved surface. Nevertheless, the total flux remains fixed by the unchanged enclosed wire charge. \(\textbf{Final answer:}\) \[ \boxed{\frac{F_f}{F_i}=\frac23,\qquad \frac{\Phi_f}{\Phi_i}=1} \]

Explanation: \(\textbf{Determine the central point charge:}\) The complete spherical surface encloses \(+Q\). The external uniform field contributes zero net flux through this closed surface. Gauss's law therefore gives \[ \Phi_{\mathrm{sphere}}=\frac{Q}{\varepsilon_0} \] The measured flux is \(8\Phi_0\): \[ \frac{Q}{\varepsilon_0}=8\Phi_0 \] Hence, \[ Q=8\varepsilon_0\Phi_0 \] \(\textbf{Find the central charge's contribution to the octant:}\) The point charge lies at the sphere's centre. Its electric field is radial and has constant magnitude over the sphere. The eight congruent spherical octants therefore receive equal flux contributions from this central source. Thus, \[ \Phi_{Q,\mathrm{octant}}=\frac18\frac{Q}{\varepsilon_0} \] Substituting, \[ \Phi_{Q,\mathrm{octant}}=\frac18(8\Phi_0) \] \[ \Phi_{Q,\mathrm{octant}}=\Phi_0 \] \(\textbf{Separate the external-field contribution:}\) The total measured octant flux is \[ \Phi_{\mathrm{octant},i}=3\Phi_0 \] The central charge contributes \(\Phi_0\). Therefore, the uniform field contributes \[ \Phi_{u,\mathrm{octant}}=3\Phi_0-\Phi_0 \] \[ \Phi_{u,\mathrm{octant}}=2\Phi_0 \] \(\textbf{Determine the projected area of the octant:}\) The uniform field points along positive \(z\). The curved spherical octant projects onto a quarter-circle in the \(xy\)-plane. Its projected area is \[ A_{\mathrm{proj}}=\frac14\pi R^2 \] For a uniform field, the octant's signed flux contribution is \[ \Phi_{u,\mathrm{octant}}=E_uA_{\mathrm{proj}} \] Thus, \[ 2\Phi_0=E_u\frac{\pi R^2}{4} \] Rearranging, \[ E_u=\frac{8\Phi_0}{\pi R^2} \] \(\textbf{Reverse the uniform field:}\) After reversal, the uniform-field contribution changes sign: \[ \Phi_{u,\mathrm{octant}}'=-2\Phi_0 \] The central point-charge contribution remains \(+\Phi_0\). Therefore, \[ \Phi_{\mathrm{octant},f}=\Phi_0-2\Phi_0 \] \[ \Phi_{\mathrm{octant},f}=-\Phi_0 \] \(\textbf{Physical check:}\) The complete sphere continues to have total outward flux \(8\Phi_0\), because reversing a uniform external field cannot change the enclosed charge. The open curved octant, however, can have a negative net flux because its external-field contribution need not cancel within that restricted surface. \(\textbf{Final answer:}\) \[ \boxed{E_u=\frac{8\Phi_0}{\pi R^2},\qquad \Phi_{\mathrm{octant},f}=-\Phi_0} \]

Explanation: \(\textbf{Determine how much sheet charge lies inside the sphere:}\) An infinite plane passing through a sphere at perpendicular distance \(d\) from its centre cuts a circular disk out of the sphere. The radius of this disk follows from the right-triangle relation \[ \rho^2+d^2=R^2 \] Therefore, \[ \rho^2=R^2-d^2 \] The sheet area lying inside the Gaussian sphere is \[ A(d)=\pi\rho^2 \] \[ A(d)=\pi(R^2-d^2) \] Only the charge on this intercepted disk is enclosed. \(\textbf{Calculate the initially enclosed sheet area:}\) Initially, \[ d_i=\frac R2 \] Thus, \[ A_i=\pi\left(R^2-\frac{R^2}{4}\right) \] \[ A_i=\frac34\pi R^2 \] The initially enclosed charge is \[ Q_{\mathrm{enc},i}=\sigma A_i \] \[ Q_{\mathrm{enc},i}=\frac34\pi\sigma R^2 \] \(\textbf{Determine the initial Gaussian flux:}\) By Gauss's law, \[ \Phi_i=\frac{Q_{\mathrm{enc},i}}{\varepsilon_0} \] \[ \Phi_i=\frac{3\pi\sigma R^2}{4\varepsilon_0} \] \(\textbf{Calculate the enclosed area after translation:}\) The new sheet position has perpendicular distance \[ d_f=\frac R{\sqrt2} \] Therefore, \[ A_f=\pi\left(R^2-\frac{R^2}{2}\right) \] \[ A_f=\frac12\pi R^2 \] The new enclosed sheet charge is \[ Q_{\mathrm{enc},f}=\frac12\pi\sigma R^2 \] Thus, \[ \Phi_f=\frac{\pi\sigma R^2}{2\varepsilon_0} \] \(\textbf{Compare the total fluxes:}\) \[ \frac{\Phi_f}{\Phi_i}=\frac{\pi\sigma R^2/(2\varepsilon_0)}{3\pi\sigma R^2/(4\varepsilon_0)} \] \[ \frac{\Phi_f}{\Phi_i}=\frac23 \] \(\textbf{Determine the force at the fixed origin:}\) The sheet is positively charged in both arrangements. The origin lies to the left of the sheet before and after translation. The field of an ideal infinite sheet has distance-independent magnitude \[ E_s=\frac{\sigma}{2\varepsilon_0} \] At the origin it points away from the sheet, along negative \(x\): \[ \vec E_O=-\frac{\sigma}{2\varepsilon_0}\hat i \] The negative test charge therefore experiences \[ \vec F_f=(-q)\vec E_O \] \[ \vec F_f=+\frac{q\sigma}{2\varepsilon_0}\hat i \] \(\textbf{Physical interpretation:}\) Translation changes the sheet area intercepted by the spherical Gaussian surface and therefore changes its enclosed charge and net flux. The field at the fixed origin does not change because the ideal infinite-sheet field depends on the side of the sheet, not the perpendicular distance. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Phi_f}{\Phi_i}=\frac23,\qquad \vec F_f=+\frac{q\sigma}{2\varepsilon_0}\hat i} \]

Explanation: \(\textbf{Identify the initially enclosed wire length:}\) The Gaussian box has vertical height \(h\). The wire makes an angle of \(60^\circ\) with the vertical axis and crosses the top and bottom faces. Let the length of the wire segment inside the box be \(\ell_i\). Its vertical projection equals the box height: \[ \ell_i\cos60^\circ=h \] Therefore, \[ \ell_i=\frac{h}{\cos60^\circ} \] \[ \ell_i=2h \] \(\textbf{Calculate the initial flux:}\) The initial enclosed charge is \[ Q_{\mathrm{enc},i}=\lambda\ell_i \] \[ Q_{\mathrm{enc},i}=2\lambda h \] Gauss's law gives \[ \Phi_i=\frac{2\lambda h}{\varepsilon_0} \] The shape of the Gaussian box does not affect this relation. \(\textbf{Apply the wire rotation and density change:}\) After rotation, the wire is vertical. Its segment inside the box has length \[ \ell_f=h \] However, its new charge density is \[ \lambda_f=2\lambda \] The final enclosed charge is \[ Q_{\mathrm{enc},f}=\lambda_f\ell_f \] \[ Q_{\mathrm{enc},f}=(2\lambda)h \] \[ Q_{\mathrm{enc},f}=2\lambda h \] Therefore, \[ \Phi_f=\frac{2\lambda h}{\varepsilon_0} \] Thus, \[ \frac{\Phi_f}{\Phi_i}=1 \] \(\textbf{Determine the initial field at P:}\) Point P is at \((0,d,0)\), directly above the \(xz\)-plane in the positive \(y\)-direction. The initial wire lies in the \(xz\)-plane and passes through the origin. Its shortest distance from P is therefore \(d\). For an infinite positive wire, \[ E_i=\frac{\lambda}{2\pi\varepsilon_0d} \] The field points along positive \(y\), away from the wire. The initial test-charge force magnitude is \[ F_i=qE_i \] \[ F_i=\frac{q\lambda}{2\pi\varepsilon_0d} \] \(\textbf{Determine the final field and force:}\) After rotation, the wire coincides with the \(z\)-axis. Point P remains at perpendicular distance \(d\). The wire's charge density has doubled: \[ E_f=\frac{2\lambda}{2\pi\varepsilon_0d} \] Therefore, \[ E_f=2E_i \] Since the test charge is unchanged, \[ \frac{F_f}{F_i}=\frac{qE_f}{qE_i} \] \[ \frac{F_f}{F_i}=2 \] \(\textbf{Physical interpretation:}\) The reduced enclosed wire length exactly compensates for the increased linear charge density, keeping the total Gaussian flux unchanged. The local field at P, however, doubles because its perpendicular distance from the wire remains \(d\) while the charge density doubles. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Phi_f}{\Phi_i}=1,\qquad \frac{F_f}{F_i}=2} \]

Explanation: Electric potential is a scalar quantity associated with the source-generated electrostatic field. It is defined as electrostatic potential energy per unit charge, using a specified reference. For a sufficiently small test charge \(q_t\), the relation is \(U=q_tV\). The potential therefore does not depend on the sign or magnitude of the test charge introduced to measure it. Using the first measurement, the potential at P is \(6U_0/(2q)=3U_0/q\). The second measurement gives \((-3U_0)/(-q)=3U_0/q\), confirming the same value. The third measurement independently gives \(9U_0/(3q)=3U_0/q\). The sign of the potential energy changes when the sign of the test charge changes, but the potential remains the same. A positive potential does not mean that a local electric-field vector must point in a particular coordinate direction. The electric field is a vector related to the spatial variation of potential, not merely to its value at a single point. Knowing the potential at only one point is insufficient to determine that variation. The local field may be nonzero or zero depending on the surrounding potential distribution. All three measurements are mutually consistent with a fixed positive potential, but none identifies the electric-field vector. Therefore, option B correctly distinguishes electric potential from both potential energy and electric field.

Explanation: \(\textbf{Apply scalar superposition at the origin:}\) The electric potential due to a point charge is \[ V=\frac{kQ_s}{r} \] where the signed source charge \(Q_s\) is used. The origin is distance \(a\) from each source. Therefore, \[ V_O=\frac{kQ}{a}-\frac{kQ}{a} \] \[ V_O=0 \] \(\textbf{Calculate the potential at P:}\) Point P is at \((0,a)\). Its distance from either source is \[ r=\sqrt{a^2+a^2} \] \[ r=\sqrt2a \] The two potential contributions are equal in magnitude and opposite in sign: \[ V_P=\frac{kQ}{\sqrt2a}-\frac{kQ}{\sqrt2a} \] \[ V_P=0 \] Thus, O and P have equal electric potential. \(\textbf{Determine the field from the positive source at P:}\) The displacement from \(+Q\) at \((-a,0)\) to P is \[ \vec r_+=a\hat i+a\hat j \] The distance is \(\sqrt2a\). The field magnitude is \[ E_+=\frac{kQ}{(\sqrt2a)^2} \] \[ E_+=\frac{E_0}{2} \] Its unit direction is \[ \hat u_+=\frac{\hat i+\hat j}{\sqrt2} \] Therefore, \[ \vec E_+=\frac{E_0}{2\sqrt2}(\hat i+\hat j) \] \(\textbf{Determine the field from the negative source at P:}\) The field due to \(-Q\) points towards that negative charge at \((a,0)\). The unit direction from P towards the source is \[ \hat u_-=\frac{\hat i-\hat j}{\sqrt2} \] The field magnitude is again \(E_0/2\). Hence, \[ \vec E_-=\frac{E_0}{2\sqrt2}(\hat i-\hat j) \] \(\textbf{Add the field vectors:}\) \[ \vec E_P=\vec E_++\vec E_- \] \[ \vec E_P=\frac{E_0}{2\sqrt2}[(\hat i+\hat j)+(\hat i-\hat j)] \] \[ \vec E_P=+\frac{E_0}{\sqrt2}\hat i \] \(\textbf{Physical distinction:}\) Potential contributions add algebraically because electric potential is a scalar. The vector fields must be added componentwise. At P, the potential contributions cancel, but the horizontal field components reinforce. Consequently, zero potential does not imply zero electric field. \(\textbf{Final answer:}\) \[ \boxed{V_O=V_P=0,\qquad \vec E_P=+\frac{E_0}{\sqrt2}\hat i} \]

Explanation: \(\textbf{Write the point-charge relations:}\) The potential of a point charge, relative to infinity, is \[ V=\frac{kQ}{r} \] The corresponding electric-field magnitude is \[ E=\frac{kQ}{r^2} \] These quantities have different distance dependences. Their ratio can eliminate the initially unknown source charge. \(\textbf{Infer the observation distance:}\) Dividing the potential relation by the field relation, \[ \frac VE=\frac{kQ/r}{kQ/r^2} \] Therefore, \[ \frac VE=r \] Substitute the measured values: \[ r=\frac{12V_0}{4E_0} \] \[ r=\frac{3V_0}{E_0} \] The result has units of length because one volt divided by one volt per metre is one metre. \(\textbf{Determine the source charge:}\) Use the measured potential: \[ V=\frac{kQ}{r} \] Rearranging, \[ Q=\frac{Vr}{k} \] Substituting \(V=12V_0\) and the inferred distance, \[ Q=\frac{12V_0}{k}\left(\frac{3V_0}{E_0}\right) \] \[ Q=\frac{36V_0^2}{kE_0} \] \(\textbf{Check against the measured field:}\) Substitute both inferred quantities into the field relation: \[ E=\frac{kQ}{r^2} \] \[ E=\frac{k[36V_0^2/(kE_0)]}{[3V_0/E_0]^2} \] \[ E=\frac{36V_0^2/E_0}{9V_0^2/E_0^2} \] \[ E=4E_0 \] This reproduces the field measurement. \(\textbf{Physical interpretation:}\) Potential and field are generated by the same point charge but decrease with different powers of distance. Combining both measurements therefore determines the distance and source strength separately. A single potential measurement would not have been sufficient to determine both unknowns. \(\textbf{Final answer:}\) \[ \boxed{r=\frac{3V_0}{E_0},\qquad Q=\frac{36V_0^2}{kE_0}} \]

Explanation: \(\textbf{Distinguish the physical potential from the reading offset:}\) For an isolated point charge, \[ V(r)=\frac{kQ}{r} \] when infinity is chosen as the zero-potential reference. The instrument adds a spatially constant offset \(C\): \[ V_{\mathrm{read}}(r)=\frac{kQ}{r}+C \] A constant potential offset does not change electric-field strength or electrostatic force. \(\textbf{Use the two measured readings:}\) At \(r=a\), \[ \frac{kQ}{a}+C=7V_0 \] At \(r=2a\), \[ \frac{kQ}{2a}+C=5V_0 \] \(\textbf{Eliminate the calibration offset:}\) Subtract the second equation from the first: \[ \frac{kQ}{a}-\frac{kQ}{2a}=2V_0 \] \[ \frac{kQ}{2a}=2V_0 \] Therefore, \[ \frac{kQ}{a}=4V_0 \] The positive result shows that the source charge is positive. \(\textbf{Determine the offset:}\) Using the first measurement, \[ 4V_0+C=7V_0 \] Therefore, \[ C=3V_0 \] \(\textbf{Predict the reading at }4a\textbf{:}\) \[ V_{\mathrm{read}}(4a)=\frac{kQ}{4a}+C \] Since \(kQ/a=4V_0\), \[ V_{\mathrm{read}}(4a)=V_0+3V_0 \] \[ V_{\mathrm{read}}(4a)=4V_0 \] \(\textbf{Determine the electric field at }r=a\textbf{:}\) The source charge is positive, so its electric field points radially outwards. Its magnitude is \[ E(a)=\frac{kQ}{a^2} \] Using \(kQ/a=4V_0\), \[ E(a)=\frac{4V_0}{a} \] Therefore, \[ \vec E(a)=+\frac{4V_0}{a}\hat r \] \(\textbf{Calculate the negative test-charge force:}\) \[ \vec F=(-q)\vec E(a) \] \[ \vec F=-\frac{4qV_0}{a}\hat r \] \(\textbf{Physical interpretation:}\) The readings themselves depend on the arbitrary instrumental potential offset. The difference between readings does not. This difference determines the physical source charge and hence the electric field. The electrostatic force is unaffected by the chosen additive potential reference. \(\textbf{Final answer:}\) \[ \boxed{V_{\mathrm{read}}(4a)=4V_0,\qquad \vec F(a)=-\frac{4qV_0}{a}\hat r} \]

Explanation: \(\textbf{Examine the original two-charge arrangement:}\) Initially, only the two equal positive charges at \(x=-a\) and \(x=+a\) are present. Their electric fields at the origin have equal magnitudes and opposite directions. Consequently, their resultant field at the origin is zero. Their electric potentials, however, are both positive and add algebraically. \(\textbf{Calculate the initial potential:}\) Each positive charge is distance \(a\) from the origin. Thus, \[ V_{O,i}=\frac{kQ}{a}+\frac{kQ}{a} \] \[ V_{O,i}=\frac{2kQ}{a} \] The original field is \[ \vec E_{O,i}=\vec0 \] This already demonstrates that zero electric field need not mean zero potential. \(\textbf{Introduce the third charge:}\) The added source is \(-2Q\), located at \((0,a)\). Its distance from the origin is \(a\). Its contribution to the potential is \[ V_3=-\frac{2kQ}{a} \] \(\textbf{Calculate the new potential at O:}\) Potential contributions are scalars and must be added algebraically: \[ V_O=V_{O,i}+V_3 \] \[ V_O=\frac{2kQ}{a}-\frac{2kQ}{a} \] \[ V_O=0 \] \(\textbf{Determine the new electric field:}\) The two original positive charges still produce zero resultant field at the origin. The negative third charge produces a field directed towards itself. Since it lies above the origin, this contribution points along positive \(y\). Its magnitude is \[ E_3=\frac{k(2Q)}{a^2} \] Therefore, \[ \vec E_3=+\frac{2kQ}{a^2}\hat j \] The final field is \[ \vec E_O=\vec E_{O,i}+\vec E_3 \] \[ \vec E_O=+\frac{2kQ}{a^2}\hat j \] \(\textbf{Physical interpretation:}\) The final charge configuration has zero total algebraic charge, but that fact does not require zero field or zero potential at a particular point. Here, the scalar potential contributions cancel. The electric-field contributions do not, because the third charge introduces an unbalanced upward field. \(\textbf{Final answer:}\) \[ \boxed{V_O=0,\qquad \vec E_O=+\frac{2kQ}{a^2}\hat j} \]

Explanation: \(\textbf{Identify the endpoint displacement:}\) The initial and final position vectors are \[ \vec r_A=\vec0 \] \[ \vec r_B=2a\hat i-a\hat j \] Therefore, \[ \Delta\vec r=\vec r_B-\vec r_A \] \[ \Delta\vec r=2a\hat i-a\hat j \] \(\textbf{Relate potential difference to the uniform field:}\) In a uniform electrostatic field, \[ V_B-V_A=-\vec E\cdot\Delta\vec r \] The negative sign indicates that potential decreases in the direction of the electric field. Substituting, \[ V_B-V_A=-E_0(3\hat i+4\hat j)\cdot(2a\hat i-a\hat j) \] \[ V_B-V_A=-E_0[(3)(2a)+(4)(-a)] \] \[ V_B-V_A=-E_0(6a-4a) \] \[ V_B-V_A=-2E_0a \] \(\textbf{Check the two displacement stages:}\) Along the horizontal stage, \[ \Delta\vec r_1=2a\hat i \] Its potential change is \[ \Delta V_1=-\vec E\cdot\Delta\vec r_1 \] \[ \Delta V_1=-6E_0a \] Along the downward stage, \[ \Delta\vec r_2=-a\hat j \] Its potential change is \[ \Delta V_2=-\vec E\cdot\Delta\vec r_2 \] \[ \Delta V_2=+4E_0a \] The total is \[ \Delta V_1+\Delta V_2=-2E_0a \] This agrees with the direct endpoint calculation. \(\textbf{Calculate electrostatic work on the negative particle:}\) The work done by the electric field in moving charge \(q_t\) from A to B is \[ W_{\mathrm{field}}=q_t(V_A-V_B) \] Here, \[ q_t=-q \] and \[ V_A-V_B=+2E_0a \] Therefore, \[ W_{\mathrm{field}}=(-q)(2E_0a) \] \[ W_{\mathrm{field}}=-2qE_0a \] \(\textbf{Physical check:}\) The field has a positive component along the net displacement, so the potential decreases from A to B. A negative charge experiences force opposite to the electric field. The electric field consequently does negative work for this displacement. Because an electrostatic field is conservative, the work is independent of the path taken between the same endpoints. \(\textbf{Final answer:}\) \[ \boxed{V_B-V_A=-2E_0a,\qquad W_{\mathrm{field}}=-2qE_0a} \]

Explanation: Electric potential due to an isolated point charge depends only on radial distance from the charge, whereas its electric field is a radial vector. Points P and R both lie at distance \(a\) from the origin. Therefore, they have the same potential, \(kQ/a\). A displacement between points of equal potential involves zero work by the electrostatic force, regardless of the particle's charge sign. Consequently, the field does zero work in transporting the negative charge from P to R. The electric field at P points radially outward along positive \(x\), while the field at R points radially outward along positive \(y\). Their magnitudes are equal because both points have the same distance from the source. Their directions are perpendicular because the two radius vectors are perpendicular. Point S lies at distance \(2a\) from the origin, so its potential is \(kQ/(2a)\). This is less than the potential at R. In moving from R to S, the particle therefore undergoes a decrease in electric potential. For a positive charge, the electrostatic field would do positive work along this potential decrease. However, the transported particle has negative charge \(-q\). Its electrostatic force acts towards the positive source, opposite to the outward displacement from radius \(a\) to radius \(2a\). Therefore, the electrostatic field does negative work during the R-to-S stage. The result distinguishes equipotential motion, radial field direction and the sign of electrostatic work for a negative charge. The source charge remains stationary throughout, but this does not require the work between different potentials to vanish.

Explanation: The Assertion is correct because the numerical zero of electric potential is a matter of reference. An electrostatic potential function can be shifted by the same additive constant everywhere without changing its physical potential differences. The electric field is related to the spatial variation of potential, rather than its absolute numerical value at one chosen point. In one dimension, for example, the electric-field component is related to the negative rate of change of potential with position. Adding a spatially constant value does not alter this rate of change. The electric-field vector therefore remains unchanged under such a reference shift. The electric force on a test charge is \(\vec F=q\vec E\), so the force also remains unchanged. The sign and magnitude of the test charge determine how it responds to the unchanged field. The electrostatic work done in moving a charge between two points depends on the potential difference between those points. That potential difference is also unaffected by adding the same constant to both endpoint potentials. Individual numerical potential values can change, but measurable electrostatic forces and work differences do not. The Reason correctly identifies this dependence on spatial potential variation and potential difference. Both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Determine the initial source distances:}\) Point P is at \(x=a\). Its distance from the charge at the origin is \[ r_1=a \] Its distance from the second charge at \(x=3a\) is \[ r_2=3a-a=2a \] The electric potential due to a point charge is \[ V=\frac{kQ_s}{r} \] where \(Q_s\) is the signed source charge. \(\textbf{Calculate the initial potential:}\) Both charges are initially positive. Their potential contributions add: \[ V_i=\frac{kQ}{a}+\frac{k(4Q)}{2a} \] \[ V_i=\frac{kQ}{a}+\frac{2kQ}{a} \] \[ V_i=\frac{3kQ}{a} \] \(\textbf{Check the initial field cancellation:}\) The first positive charge produces a field towards positive \(x\): \[ \vec E_1=+E_0\hat i \] The second positive charge produces a field towards negative \(x\): \[ \vec E_{2,i}=-\frac{k(4Q)}{(2a)^2}\hat i \] \[ \vec E_{2,i}=-E_0\hat i \] Thus, \[ \vec E_i=\vec0 \] The given zero-field observation is consistent with the geometry. \(\textbf{Reverse the second source charge:}\) After the change, the second charge is \(-4Q\). Its potential contribution becomes negative: \[ V_f=\frac{kQ}{a}-\frac{k(4Q)}{2a} \] \[ V_f=\frac{kQ}{a}-\frac{2kQ}{a} \] \[ V_f=-\frac{kQ}{a} \] \(\textbf{Recalculate the field direction:}\) The field due to \(+Q\) remains \(+E_0\hat i\). The negative second charge attracts a positive test charge towards \(x=3a\), so its field at P also points towards positive \(x\): \[ \vec E_{2,f}=+E_0\hat i \] Consequently, \[ \vec E_f=E_0\hat i+E_0\hat i \] \[ \vec E_f=+2E_0\hat i \] \(\textbf{Determine the test-charge force:}\) \[ \vec F_f=q\vec E_f \] \[ \vec F_f=+2qE_0\hat i \] \(\textbf{Physical interpretation:}\) Initially, the field contributions cancel while the positive scalar potential contributions reinforce. Reversing the second charge reverses both its potential sign and field direction. The final potential becomes negative, although the resultant field points towards positive \(x\). \(\textbf{Final answer:}\) \[ \boxed{V_i=\frac{3kQ}{a},\qquad V_f=-\frac{kQ}{a},\qquad \vec F_f=+2qE_0\hat i} \]

Explanation: \(\textbf{Identify the two potential contributions:}\) The arrangement contains a uniformly charged spherical shell and a negative point charge at its centre. The total electric potential is the algebraic sum of their potentials. Because both source distributions are centred at the same point, the resultant field is spherically symmetric. However, the potential and field must be evaluated separately in the interior and exterior regions. \(\textbf{Potential due to the shell inside it:}\) The shell radius is \(2a\). At every point strictly inside a uniformly charged thin spherical shell, the shell's electric field is zero. Its potential is constant throughout the interior and equals its surface potential: \[ V_{\mathrm{shell}}(r\lt2a)=\frac{kQ}{2a} \] \(\textbf{Calculate the potential at P:}\) Point P lies at radius \(a\), inside the shell. The central point charge contributes \[ V_{\mathrm{centre}}(P)=-\frac{kQ}{a} \] Therefore, \[ V_P=\frac{kQ}{2a}-\frac{kQ}{a} \] \[ V_P=-\frac{kQ}{2a} \] \(\textbf{Calculate the potential at R:}\) Point R lies outside the shell because \[ 3a\gt2a \] The shell's exterior potential is \[ V_{\mathrm{shell}}(R)=+\frac{kQ}{3a} \] The central charge contributes \[ V_{\mathrm{centre}}(R)=-\frac{kQ}{3a} \] Thus, \[ V_R=\frac{kQ}{3a}-\frac{kQ}{3a} \] \[ V_R=0 \] \(\textbf{Determine the electric field at P:}\) The uniformly charged shell produces zero electric field at P. The central negative charge produces a radially inward field of magnitude \[ E_{\mathrm{centre}}(P)=\frac{kQ}{a^2} \] Therefore, \[ \vec E_P=-\frac{kQ}{a^2}\hat r \] \(\textbf{Check the exterior behaviour:}\) Outside the shell, the total algebraic enclosed charge is \[ Q_{\mathrm{total}}=+Q-Q=0 \] Spherical symmetry consequently gives zero resultant exterior field. The exterior potential is also zero under the infinity reference. Inside the shell, however, the central negative charge produces a nonzero field. \(\textbf{Physical distinction:}\) The spherical shell's zero interior field does not mean that another source placed inside it cannot produce a field there. The shell contributes a constant interior potential, which must still be included in the scalar superposition. \(\textbf{Final answer:}\) \[ \boxed{V_P=-\frac{kQ}{2a},\qquad V_R=0,\qquad \vec E_P=-\frac{kQ}{a^2}\hat r} \]

Explanation: \(\textbf{Separate the potential regions:}\) The potential due to a uniformly charged spherical shell is constant throughout its interior. At points outside that shell, its potential is the same as that of its total charge concentrated at the centre. Potential contributions from the two shells add algebraically. \(\textbf{Calculate the potential at P:}\) Point P lies between the shells: \[ R\lt\frac{3R}{2}\lt2R \] It is outside the inner shell but inside the outer shell. The inner-shell potential at P is \[ V_1(P)=\frac{kQ}{3R/2} \] \[ V_1(P)=\frac{2kQ}{3R} \] The outer-shell potential is its constant interior value: \[ V_2(P)=\frac{k(-2Q)}{2R} \] \[ V_2(P)=-\frac{kQ}{R} \] Therefore, \[ V_P=\frac{2kQ}{3R}-\frac{kQ}{R} \] \[ V_P=-\frac{kQ}{3R} \] \(\textbf{Calculate the potential at S:}\) Point S lies outside both shells. Their combined charge is \[ Q_{\mathrm{total}}=Q-2Q=-Q \] Thus, \[ V_S=\frac{k(-Q)}{3R} \] \[ V_S=-\frac{kQ}{3R} \] \(\textbf{Determine the work done by the electrostatic field:}\) The field work in moving charge \(+q\) between two points is \[ W_{P\to S}=q(V_P-V_S) \] Since the two potentials are equal, \[ W_{P\to S}=q\left(-\frac{kQ}{3R}+\frac{kQ}{3R}\right) \] \[ W_{P\to S}=0 \] \(\textbf{Compare the local electric fields:}\) At P, a concentric Gaussian surface encloses only \(+Q\). Therefore, \[ \vec E_P=+\frac{kQ}{(3R/2)^2}\hat r \] \[ \vec E_P=+\frac{4kQ}{9R^2}\hat r \] At S, both shells are enclosed, giving total charge \(-Q\): \[ \vec E_S=-\frac{kQ}{(3R)^2}\hat r \] \[ \vec E_S=-\frac{kQ}{9R^2}\hat r \] The fields are nonzero and point in opposite radial directions. \(\textbf{Physical interpretation:}\) Equal endpoint potentials imply zero net electrostatic work, even when the electric field is nonzero along portions of the path. Here the radial field changes direction between the two shell regions. The positive and negative work contributions cancel over the complete displacement. \(\textbf{Final answer:}\) The potentials at P and S are equal and negative, the net electrostatic work is zero, and the local fields point outwards at P and inwards at S.

Explanation: \(\textbf{Distinguish the shell field from the external field:}\) Both endpoints lie strictly inside the uniformly charged spherical shell. The shell produces zero electric field at every interior point. Its electric potential throughout the interior is a spatially constant value. Changing the sign of the shell charge changes that constant potential contribution but does not introduce an interior field. \(\textbf{Determine the total interior electric field:}\) The applied uniform field is \[ \vec E_u=E_0\hat i \] The shell field is zero inside: \[ \vec E_{\mathrm{shell}}=\vec0 \] Consequently, the total interior field is \[ \vec E_{\mathrm{total}}=E_0\hat i \] This remains unchanged when the shell charge is reversed. \(\textbf{Determine the displacement between endpoints:}\) The initial point is \[ \vec r_P=-\frac R2\hat i \] The final point is \[ \vec r_S=+\frac R2\hat i \] Thus, \[ \Delta\vec r=\vec r_S-\vec r_P \] \[ \Delta\vec r=R\hat i \] \(\textbf{Calculate the potential difference:}\) For a uniform electric field, \[ V_S-V_P=-\vec E\cdot\Delta\vec r \] Substituting, \[ V_S-V_P=-(E_0\hat i)\cdot(R\hat i) \] \[ V_S-V_P=-E_0R \] The shell's constant interior potential cancels when the endpoint potentials are subtracted. \(\textbf{Calculate the electrostatic work:}\) The test charge is \[ q_t=-q \] The work done by the electric field is \[ W_{\mathrm{field}}=q_t(V_P-V_S) \] Since \[ V_P-V_S=+E_0R \] we obtain \[ W_{\mathrm{field}}=(-q)(E_0R) \] \[ W_{\mathrm{field}}=-qE_0R \] \(\textbf{Apply the shell-charge reversal:}\) After the shell charge becomes \(-Q\), its interior field is still zero. The total field remains \(E_0\hat i\). Therefore, the potential difference and electric work remain exactly the same. \(\textbf{Physical interpretation:}\) A change in the absolute potential contribution of a source does not necessarily change potential differences. The spherical shell changes the common potential at both interior points, while the external uniform field determines their potential difference. \(\textbf{Final answer:}\) \[ \boxed{V_S-V_P=-E_0R,\qquad W_{\mathrm{field}}=-qE_0R\quad\text{in both experiments}} \]

Explanation: \(\textbf{Identify the governing shell-potential behaviour:}\) For a uniformly charged thin spherical shell of radius \(b\), the potential relative to infinity is \[ V(r)=\frac{kQ}{b},\qquad r\leq b \] and \[ V(r)=\frac{kQ}{r},\qquad r\geq b \] The potential is continuous at the surface. Unlike the electric field, it generally remains nonzero inside a charged shell. \(\textbf{Calculate the initial potential at P:}\) Initially, the shell radius is \(R\). Point P lies at \[ r_P=\frac R2 \] Thus, P is inside the shell: \[ V_P^{(i)}=\frac{kQ}{R} \] \(\textbf{Calculate the initial potential at S:}\) Point S lies at \(r_S=3R\), outside the original shell. Therefore, \[ V_S^{(i)}=\frac{kQ}{3R} \] \(\textbf{Expand the shell at fixed total charge:}\) The new shell radius is \[ R'=2R \] The total shell charge remains \(+Q\). Consequently, its surface charge density decreases, but its total charge does not. \(\textbf{Re-evaluate P after expansion:}\) The point P remains at radius \(R/2\). It lies inside the expanded shell as well. The constant interior potential is now \[ V_P^{(f)}=\frac{kQ}{2R} \] Therefore, \[ \frac{V_P^{(f)}}{V_P^{(i)}}=\frac{kQ/(2R)}{kQ/R} \] \[ \frac{V_P^{(f)}}{V_P^{(i)}}=\frac12 \] \(\textbf{Re-evaluate S after expansion:}\) Point S remains at radius \(3R\). It is still outside the expanded shell because \[ 3R\gt2R \] Its potential is therefore \[ V_S^{(f)}=\frac{kQ}{3R} \] Consequently, \[ \frac{V_S^{(f)}}{V_S^{(i)}}=1 \] \(\textbf{Physical distinction:}\) The potential inside the shell is spatially constant but depends on the shell radius at fixed charge. The exterior potential at a fixed observation point depends only on the total shell charge and the point's distance from the centre. Thus, shell expansion changes the interior potential while leaving the exterior potential unchanged at S. \(\textbf{Final answer:}\) \[ \boxed{\frac{V_P^{(f)}}{V_P^{(i)}}=\frac12,\qquad \frac{V_S^{(f)}}{V_S^{(i)}}=1} \]

Explanation: \(\textbf{Determine the two axial distances at P:}\) At \(P(2a,0,0)\), the distance from the positive source at \(x=a\) is \[ r_+=2a-a=a \] The distance from the negative source at \(x=-a\) is \[ r_-=2a-(-a)=3a \] \(\textbf{Calculate the exact potential at P:}\) Electric potential is a scalar. The signed contributions give \[ V_P=\frac{kQ}{a}-\frac{kQ}{3a} \] \[ V_P=+\frac{2kQ}{3a} \] \(\textbf{Calculate the exact potential at S:}\) At \(S(-2a,0,0)\), the negative source is nearer, at distance \(a\). The positive source is at distance \(3a\). Therefore, \[ V_S=-\frac{kQ}{a}+\frac{kQ}{3a} \] \[ V_S=-\frac{2kQ}{3a} \] The two exterior axial potentials have equal magnitudes and opposite signs. \(\textbf{Determine the field at P:}\) The positive source produces a field towards positive \(x\): \[ \vec E_+(P)=+\frac{kQ}{a^2}\hat i \] The negative source produces a field towards itself, along negative \(x\): \[ \vec E_-(P)=-\frac{kQ}{(3a)^2}\hat i \] Thus, \[ \vec E_P=\left(\frac{kQ}{a^2}-\frac{kQ}{9a^2}\right)\hat i \] \[ \vec E_P=+\frac{8kQ}{9a^2}\hat i \] \(\textbf{Determine the field at S:}\) The negative source at \(x=-a\) produces a field directed towards positive \(x\), because S lies to its left. The positive source at \(x=a\) produces a weaker field towards negative \(x\). Therefore, \[ \vec E_S=\left(\frac{kQ}{a^2}-\frac{kQ}{9a^2}\right)\hat i \] \[ \vec E_S=+\frac{8kQ}{9a^2}\hat i \] \(\textbf{Interpret the symmetry:}\) The two points are related by reflection about the dipole midpoint. Their scalar potentials reverse sign, but their electric-field vectors remain identical. This is a distinctive symmetry property of the axial field and potential of an electric dipole. \(\textbf{Final answer:}\) \[ \boxed{V_P=+\frac{2kQ}{3a},\quad V_S=-\frac{2kQ}{3a},\quad \vec E_P=\vec E_S=+\frac{8kQ}{9a^2}\hat i} \]

Explanation: \(\textbf{Identify the equatorial geometry:}\) The dipole axis lies along the \(x\)-axis. Both P and S lie on its perpendicular bisector. At either point, the distances from the positive and negative dipole charges are equal. Consequently, the scalar potential contributions from the dipole cancel at both points: \[ V_{\mathrm{dipole}}(P)=V_{\mathrm{dipole}}(S)=0 \] The measured potential difference therefore comes entirely from the uniform external field. \(\textbf{Determine the uniform field:}\) The displacement from S to P is \[ \Delta\vec r=\vec r_P-\vec r_S \] \[ \Delta\vec r=2a\hat j \] For a uniform electric field, \[ V_P-V_S=-\vec E_u\cdot\Delta\vec r \] Since \(\vec E_u=E_u\hat j\), \[ -4V_0=-(E_u\hat j)\cdot(2a\hat j) \] \[ -4V_0=-2aE_u \] Therefore, \[ E_u=\frac{2V_0}{a} \] \(\textbf{Calculate the dipole field at P:}\) At \(P(0,a)\), the distance from either dipole charge is \[ r=\sqrt{a^2+a^2}=\sqrt2a \] The field magnitude contributed by either charge is \[ E_1=\frac{kQ}{r^2} \] \[ E_1=\frac{kQ}{2a^2} \] \(\textbf{Resolve the two dipole contributions:}\) The field due to \(+Q\) points away from \((a,0)\): \[ \vec E_+=\frac{kQ}{2\sqrt2a^2}(-\hat i+\hat j) \] The field due to \(-Q\) points towards \((-a,0)\): \[ \vec E_-=\frac{kQ}{2\sqrt2a^2}(-\hat i-\hat j) \] The vertical components cancel, while the horizontal components add: \[ \vec E_d(P)=\vec E_++\vec E_- \] \[ \vec E_d(P)=-\frac{kQ}{\sqrt2a^2}\hat i \] \(\textbf{Combine the dipole and uniform fields:}\) \[ \vec E_{\mathrm{total}}(P)=\vec E_d(P)+\vec E_u \] \[ \vec E_{\mathrm{total}}(P)=-\frac{kQ}{\sqrt2a^2}\hat i+\frac{2V_0}{a}\hat j \] \(\textbf{Determine the test-charge force:}\) The test charge is positive: \[ \vec F=q\vec E_{\mathrm{total}}(P) \] Therefore, \[ \vec F=q\left(-\frac{kQ}{\sqrt2a^2}\hat i+\frac{2V_0}{a}\hat j\right) \] \(\textbf{Physical interpretation:}\) The dipole contributes zero potential at both observation points but produces a nonzero horizontal electric field. The uniform field produces the measured potential difference and adds a vertical field component. Scalar potential cancellation therefore does not imply cancellation of the vector field. \(\textbf{Final answer:}\) \[ \boxed{E_u=\frac{2V_0}{a},\qquad \vec F=q\left(-\frac{kQ}{\sqrt2a^2}\hat i+\frac{2V_0}{a}\hat j\right)} \]