Electrostatics MCQs with Answers – Part 5 (NEET Physics)

Explanation: \(\textbf{Relate dipole energy and torque to orientation:}\) The potential energy of a dipole in a uniform electric field is \[ U=-pE\cos\theta \] The magnitude of its electric torque is \[ \tau=pE\sin\theta \] The measured energy is negative, consistent with an acute angle between the moment and field. \(\textbf{Use the energy measurement:}\) \[ -pE\cos\theta=-3U_0 \] Therefore, \[ pE\cos\theta=3U_0 \] \(\textbf{Use the torque measurement:}\) The angle is acute, so the sine is positive. Thus, \[ pE\sin\theta=4U_0 \] \(\textbf{Eliminate the unknown angle:}\) Squaring and adding the two relations, \[ (pE)^2(\cos^2\theta+\sin^2\theta)=9U_0^2+16U_0^2 \] Using \[ \cos^2\theta+\sin^2\theta=1 \] we obtain \[ (pE)^2=25U_0^2 \] Since \(pE\gt0\), \[ pE=5U_0 \] \(\textbf{Identify the initial and final energies:}\) The initial energy is \[ U_i=-3U_0 \] At \(90^\circ\), \[ U_f=-pE\cos90^\circ \] \[ U_f=0 \] \(\textbf{Determine the rotational field work:}\) For a dipole rotating in a fixed uniform field, \[ W_{\mathrm{field}}=U_i-U_f \] Therefore, \[ W_{\mathrm{field}}=-3U_0-0 \] \[ W_{\mathrm{field}}=-3U_0 \] \(\textbf{Physical interpretation:}\) The energy and torque measurements supply two independent angular projections of \(pE\). Rotating the dipole from an acute orientation towards \(90^\circ\) moves it away from its preferred parallel alignment. The electric torque opposes the imposed rotation, so the field does negative work. \(\textbf{Final answer:}\) \[ \boxed{pE=5U_0,\qquad W_{\mathrm{field}}=-3U_0} \]

Explanation: Statement I is correct because the potential energy of a dipole in a uniform field is \(U=-pE\cos\theta\). At parallel alignment, \(\theta=0^\circ\), so the energy reaches its minimum value \(-pE\). The torque magnitude \(pE|\sin\theta|\) is zero at this orientation. A small angular displacement raises the potential energy, and the electric torque tends to restore the dipole towards parallel alignment. Statement II is correct because antiparallel alignment corresponds to \(\theta=180^\circ\). The potential energy then reaches its maximum value \(+pE\), while the torque magnitude is again zero. Statement III is incorrect because zero torque at one exact orientation establishes equilibrium but does not by itself establish stability. A small displacement away from the antiparallel direction lowers potential energy. The resulting torque drives the dipole farther from the antiparallel orientation, making that equilibrium unstable. Statement IV is correct because at \(90^\circ\), the cosine is zero and the potential energy under the stated convention is zero. The sine has unit magnitude at this angle, so the electric torque has its maximum magnitude \(pE\). The dipole experiences no net translational force in an ideal uniform field, but it may experience a rotational torque. The stability distinction requires examining the change in potential energy or the direction of torque after a small angular disturbance. Therefore, Statements I, II and IV are correct.

Explanation: \(\textbf{Use the scalar product for potential energy:}\) The dipole potential energy in a uniform field is \[ U=-\vec p\cdot\vec E \] Substituting, \[ U_i=-\frac{p_0E_0}{5}(4\hat i-3\hat j)\cdot(3\hat i+4\hat j) \] \[ U_i=-\frac{p_0E_0}{5}(12-12) \] \[ U_i=0 \] The moment is perpendicular to the applied field. \(\textbf{Determine the initial torque direction:}\) The torque is a vector cross product: \[ \vec\tau_i=\vec p\times\vec E \] Substituting the given components, \[ \vec\tau_i=\frac{p_0E_0}{5}(4\hat i-3\hat j)\times(3\hat i+4\hat j) \] Only the cross terms contribute: \[ \vec\tau_i=\frac{p_0E_0}{5}[16(\hat i\times\hat j)-9(\hat j\times\hat i)] \] Using \[ \hat i\times\hat j=\hat k,\qquad \hat j\times\hat i=-\hat k \] we obtain \[ \vec\tau_i=\frac{p_0E_0}{5}(16+9)\hat k \] \[ \vec\tau_i=+5p_0E_0\hat k \] \(\textbf{Apply the dipole reversal:}\) After rotation through \(180^\circ\), \[ \vec p_f=-\vec p \] The final potential energy is \[ U_f=-\vec p_f\cdot\vec E \] Since the original scalar product was zero, \[ U_f=0 \] \(\textbf{Determine the final torque:}\) \[ \vec\tau_f=\vec p_f\times\vec E \] \[ \vec\tau_f=(-\vec p)\times\vec E \] \[ \vec\tau_f=-\vec\tau_i \] Therefore, \[ \vec\tau_f=-5p_0E_0\hat k \] \(\textbf{Physical interpretation:}\) Perpendicular alignment produces zero potential energy under the standard dipole-energy convention, but maximum torque magnitude. Reversing the moment preserves perpendicularity and therefore preserves zero energy. It reverses the cross-product direction, making the torque vector reverse. \(\textbf{Final answer:}\) \[ \boxed{U_i=U_f=0,\qquad \vec\tau_i=+5p_0E_0\hat k,\qquad \vec\tau_f=-5p_0E_0\hat k} \]

Explanation: \(\textbf{Identify the initial moment vector:}\) Initially, the dipole moment is \[ \vec p_i=p\hat i \] The uniform electric field is \[ \vec E=E_0(3\hat i+4\hat j) \] \(\textbf{Calculate the initial potential energy:}\) The energy is \[ U=-\vec p\cdot\vec E \] Therefore, \[ U_i=-(p\hat i)\cdot E_0(3\hat i+4\hat j) \] \[ U_i=-3pE_0 \] \(\textbf{Represent the rotated dipole moment:}\) A clockwise rotation through \(90^\circ\) sends positive \(x\) into negative \(y\). Thus, \[ \vec p_f=-p\hat j \] The moment magnitude remains \(p\). \(\textbf{Calculate the final potential energy:}\) \[ U_f=-\vec p_f\cdot\vec E \] Substituting, \[ U_f=-(-p\hat j)\cdot E_0(3\hat i+4\hat j) \] \[ U_f=+4pE_0 \] \(\textbf{Determine the energy change:}\) \[ \Delta U=U_f-U_i \] \[ \Delta U=4pE_0-(-3pE_0) \] \[ \Delta U=+7pE_0 \] \(\textbf{Determine the field work:}\) The external field remains fixed throughout the rotation. Its torque is conservative with potential energy \(U=-\vec p\cdot\vec E\). Therefore, \[ W_{\mathrm{field}}=-\Delta U \] \[ W_{\mathrm{field}}=-7pE_0 \] \(\textbf{Physical interpretation:}\) The initial moment has a positive projection along the field. After clockwise rotation, its projection is negative. The dipole is therefore moved from a negative-energy orientation to a positive-energy orientation. The electric field opposes this imposed rotation overall and does negative work. \(\textbf{Final answer:}\) \[ \boxed{\Delta U=+7pE_0,\qquad W_{\mathrm{field}}=-7pE_0} \]

Explanation: \(\textbf{Use both angular measurements:}\) The potential energy is \[ U=-pE\cos\theta \] The torque magnitude is \[ \tau=pE\sin\theta \] The specified angle lies in the second quadrant, so its cosine is negative and its sine is positive. \(\textbf{Apply the energy measurement:}\) \[ -pE\cos\theta=3U_0 \] Therefore, \[ pE\cos\theta=-3U_0 \] \(\textbf{Apply the torque measurement:}\) \[ pE\sin\theta=4U_0 \] \(\textbf{Eliminate the unknown angle:}\) Square and add: \[ (pE)^2(\cos^2\theta+\sin^2\theta)=9U_0^2+16U_0^2 \] Hence, \[ (pE)^2=25U_0^2 \] \[ pE=5U_0 \] \(\textbf{Interpret the initial orientation:}\) The angular relations give \[ \cos\theta=-\frac35,\qquad \sin\theta=\frac45 \] The negative cosine agrees with the given obtuse orientation. \(\textbf{Calculate the final energy:}\) Antiparallel alignment corresponds to \[ \theta_f=180^\circ \] Its potential energy is \[ U_f=-pE\cos180^\circ \] \[ U_f=+pE=5U_0 \] \(\textbf{Determine the quasistatic external work:}\) The initial energy is \[ U_i=3U_0 \] Therefore, \[ \Delta U=U_f-U_i \] \[ \Delta U=5U_0-3U_0 \] \[ \Delta U=+2U_0 \] For slow rotation without kinetic-energy change, \[ W_{\mathrm{ext}}=\Delta U=+2U_0 \] \(\textbf{Physical interpretation:}\) The dipole is rotated towards an antiparallel orientation of higher potential energy. The external agent must work against the field's tendency to turn the dipole towards parallel alignment. The torque measurement and energy measurement together determine the otherwise unknown energy scale \(pE\). \(\textbf{Final answer:}\) \[ \boxed{pE=5U_0,\qquad W_{\mathrm{ext}}=+2U_0} \]

Explanation: \(\textbf{Identify the rotational energy conversion:}\) A dipole in a uniform field has potential energy \[ U(\theta)=-pE\cos\theta \] The field exerts a torque tending to align the moment with the field. After release, the dipole rotates towards smaller angles. Because the pivot is fixed and friction is neglected, its decrease in electrostatic potential energy becomes rotational kinetic energy. \(\textbf{Calculate the initial potential energy:}\) Initially, \[ \theta_i=150^\circ \] Using \[ \cos150^\circ=-\frac{\sqrt3}{2} \] we obtain \[ U_i=-pE\left(-\frac{\sqrt3}{2}\right) \] \[ U_i=+\frac{\sqrt3}{2}pE \] \(\textbf{Calculate the potential energy at }30^\circ\textbf{:}\) At the specified later orientation, \[ \theta_f=30^\circ \] Using \[ \cos30^\circ=\frac{\sqrt3}{2} \] the potential energy is \[ U_f=-\frac{\sqrt3}{2}pE \] \(\textbf{Determine the decrease in potential energy:}\) \[ U_i-U_f=\frac{\sqrt3}{2}pE-\left(-\frac{\sqrt3}{2}pE\right) \] \[ U_i-U_f=\sqrt3pE \] \(\textbf{Apply conservation of mechanical energy:}\) The dipole starts from rest. Therefore, its initial rotational kinetic energy is zero. At the final orientation, \[ K_{\mathrm{rot}}=\frac12I\omega^2 \] Conservation of energy gives \[ \frac12I\omega^2=\sqrt3pE \] Thus, \[ \omega^2=\frac{2\sqrt3pE}{I} \] Taking the positive angular-speed value, \[ \omega=\sqrt{\frac{2\sqrt3pE}{I}} \] \(\textbf{Physical check:}\) The initial obtuse orientation has positive potential energy. The later acute orientation has negative potential energy. Their difference is available as rotational kinetic energy, making the angular speed real and nonzero. \(\textbf{Final answer:}\) \[ \boxed{\omega=\sqrt{\frac{2\sqrt3pE}{I}}} \]

Explanation: \(\textbf{Determine the initial dipole moment:}\) The electric dipole moment points from the negative charge towards the positive charge. Initially, the negative charge is at \(x=+a\), while the positive charge is at \(x=-a\). Therefore, the moment points along negative \(x\): \[ \vec p_i=-2qa\hat i \] Its magnitude is \[ p=2qa \] \(\textbf{Calculate the initial torque:}\) The applied electric field is \[ \vec E=E\hat j \] The torque is \[ \vec\tau_i=\vec p_i\times\vec E \] Substituting, \[ \vec\tau_i=(-2qa\hat i)\times(E\hat j) \] \[ \vec\tau_i=-2qaE\hat k \] The electric torque therefore tends to rotate the dipole clockwise. \(\textbf{Identify the imposed rotation direction:}\) The dipole is instead rotated anticlockwise through \(90^\circ\). An anticlockwise rotation sends the initial negative-\(x\) moment direction into negative \(y\). Thus, \[ \vec p_f=-2qa\hat j \] The final moment is antiparallel to the external field. \(\textbf{Calculate the initial energy:}\) \[ U_i=-\vec p_i\cdot\vec E \] \[ U_i=-(-2qa\hat i)\cdot(E\hat j) \] \[ U_i=0 \] The initial moment and field are perpendicular. \(\textbf{Calculate the final energy:}\) \[ U_f=-\vec p_f\cdot\vec E \] \[ U_f=-(-2qa\hat j)\cdot(E\hat j) \] \[ U_f=+2qaE \] \(\textbf{Determine the field work:}\) \[ W_{\mathrm{field}}=U_i-U_f \] \[ W_{\mathrm{field}}=0-2qaE \] \[ W_{\mathrm{field}}=-2qaE \] \(\textbf{Physical interpretation:}\) The initial torque points along negative \(z\), corresponding to clockwise rotation. The imposed anticlockwise rotation moves the dipole against this torque towards its maximum-energy orientation. Consequently, the electric field does negative work. \(\textbf{Final answer:}\) \[ \boxed{\vec\tau_i=-2qaE\hat k,\qquad W_{\mathrm{field}}=-2qaE} \]

Explanation: The dipole potential energy is \(U=-pE\cos\theta\). At P, the moment is parallel to the field, and the energy has its minimum value \(-pE\). At Q, the angle is \(60^\circ\), giving energy \(-pE/2\). At R, the angle is \(120^\circ\), giving energy \(+pE/2\). At S, the moment is antiparallel to the field, and the energy reaches its maximum value \(+pE\). Therefore, the potential energy increases successively through P, Q, R and S. The torque magnitude is \(pE|\sin\theta|\). Its magnitude is zero at P and S because the sine vanishes at both parallel and antiparallel alignment. At Q and R, the sine magnitudes are equal, so their torque magnitudes are equal and nonzero. Their equal torque magnitudes do not imply equal potential energies. The potential-energy increase during rotation from P to S is \(2pE\). An external agent performing the rotation quasistatically must supply this positive work against the field's aligning tendency. The electric field itself performs negative work during the same rotation. The parallel state is stable, while the antiparallel state is unstable, even though both have zero torque at the exact equilibrium orientations. Only option A correctly combines the energy ordering, torque comparison and work sign.

Explanation: \(\textbf{Determine the two initial energies:}\) Each dipole has potential energy \[ U=-pE\cos\theta \] For A, \[ U_{A,i}=-pE\cos60^\circ \] \[ U_{A,i}=-\frac12pE \] For B, \[ U_{B,i}=-pE\cos120^\circ \] \[ U_{B,i}=+\frac12pE \] Therefore, \[ U_i=U_{A,i}+U_{B,i} \] \[ U_i=0 \] \(\textbf{Apply the rotation of dipole A:}\) Rotating A through \(180^\circ\) reverses its moment. Its new angle measured anticlockwise is \(240^\circ\). Therefore, \[ U_{A,f}=-pE\cos240^\circ \] \[ U_{A,f}=+\frac12pE \] Dipole B remains at \(120^\circ\), so \[ U_{B,f}=+\frac12pE \] Consequently, \[ U_f=+pE \] \(\textbf{Determine the initial torque of A:}\) For a moment in the \(xy\)-plane, \[ \vec\tau=\vec p\times(E\hat i) \] The \(z\)-component is \[ \tau_z=-pE\sin\theta \] Thus, \[ \vec\tau_{A,i}=-pE\sin60^\circ\,\hat k \] \[ \vec\tau_{A,i}=-\frac{\sqrt3}{2}pE\hat k \] \(\textbf{Determine the final torque of A:}\) At \(240^\circ\), \[ \sin240^\circ=-\frac{\sqrt3}{2} \] Therefore, \[ \vec\tau_{A,f}=+\frac{\sqrt3}{2}pE\hat k \] The moment reversal has reversed A's torque. \(\textbf{Determine the change in total torque:}\) Dipole B has not moved, so its torque contribution is unchanged. Therefore, \[ \Delta\vec\tau=\vec\tau_{A,f}-\vec\tau_{A,i} \] \[ \Delta\vec\tau=\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right)pE\hat k \] \[ \Delta\vec\tau=+\sqrt3pE\hat k \] \(\textbf{Physical interpretation:}\) The initial energies cancel, but the dipoles' initial torques do not have to cancel. Reversing only A changes its energy sign and torque direction while leaving B unchanged. The total energy and total torque must therefore be tracked by superposing the separate contributions. \(\textbf{Final answer:}\) \[ \boxed{U_i=0,\qquad U_f=+pE,\qquad \Delta\vec\tau=+\sqrt3pE\hat k} \]

Explanation: \(\textbf{Calculate the initial dipole energy:}\) For a dipole in a uniform electric field, \[ U=-pE\cos\theta \] Initially, the angle is \(0^\circ\) and the field magnitude is \(E_0\): \[ U_i=-pE_0 \] \(\textbf{Calculate the first rotational work:}\) After rotation to \(60^\circ\) in the original field, \[ U_1=-pE_0\cos60^\circ \] \[ U_1=-\frac12pE_0 \] For quasistatic rotation at fixed field, \[ W_{\mathrm{rot},1}=U_1-U_i \] \[ W_{\mathrm{rot},1}=-\frac12pE_0+pE_0 \] \[ W_{\mathrm{rot},1}=+\frac12pE_0 \] \(\textbf{Apply the intermediate field-strength change:}\) The dipole remains fixed at \(60^\circ\). The field magnitude doubles: \[ E'=2E_0 \] The new dipole energy is \[ U_2=-p(2E_0)\cos60^\circ \] \[ U_2=-pE_0 \] Thus, the energy changes during this stage by \[ \Delta U_{\mathrm{field}}=U_2-U_1 \] \[ \Delta U_{\mathrm{field}}=-\frac12pE_0 \] No rotational displacement occurs, so this energy change is not rotational work done by the turning agent. \(\textbf{Calculate the final rotational work:}\) The dipole is rotated from \(60^\circ\) to \(120^\circ\) while the stronger field remains fixed. Its final energy is \[ U_f=-p(2E_0)\cos120^\circ \] \[ U_f=+pE_0 \] Therefore, \[ W_{\mathrm{rot},3}=U_f-U_2 \] \[ W_{\mathrm{rot},3}=pE_0-(-pE_0) \] \[ W_{\mathrm{rot},3}=+2pE_0 \] \(\textbf{Add the rotational works:}\) \[ W_{\mathrm{rot,total}}=\frac12pE_0+2pE_0 \] \[ W_{\mathrm{rot,total}}=\frac52pE_0 \] \(\textbf{Compare with the complete energy change:}\) \[ \Delta U_{\mathrm{overall}}=U_f-U_i \] \[ \Delta U_{\mathrm{overall}}=pE_0-(-pE_0) \] \[ \Delta U_{\mathrm{overall}}=2pE_0 \] \(\textbf{Physical interpretation:}\) The sum of the rotational works differs from the overall energy change because the field strength is altered during the intermediate stage. The dipole's energy changes when the field source is adjusted even though the dipole orientation remains fixed. The complete energy accounting includes that intermediate change. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{rot,total}}=\frac52pE_0,\qquad \Delta U_{\mathrm{overall}}=2pE_0} \]

Explanation: \(\textbf{Extract the original dipole-energy scale:}\) The dipole potential energy is \[ U=-pE\cos\theta \] At parallel alignment, \[ U(0^\circ)=-pE \] The table gives \[ -pE=-6U_0 \] Therefore, \[ pE=6U_0 \] The entries at \(90^\circ\) and \(180^\circ\) are consistent with this relation. \(\textbf{Apply the reduced field magnitude:}\) The dipole moment remains unchanged. The new field magnitude is \[ E'=\frac E3 \] Thus, \[ pE'=\frac{pE}{3} \] \[ pE'=2U_0 \] \(\textbf{Determine the initial torque in the reduced field:}\) The electric torque magnitude is \[ \tau=pE'|\sin\theta| \] At \(120^\circ\), \[ \tau_{120}=2U_0\sin120^\circ \] Using \[ \sin120^\circ=\frac{\sqrt3}{2} \] we obtain \[ \tau_{120}=\sqrt3U_0 \] \(\textbf{Calculate the initial potential energy:}\) The dipole starts the rotation at \(120^\circ\). Therefore, \[ U_i=-pE'\cos120^\circ \] \[ U_i=-2U_0\left(-\frac12\right) \] \[ U_i=+U_0 \] \(\textbf{Calculate the final potential energy:}\) At \(30^\circ\), \[ U_f=-pE'\cos30^\circ \] \[ U_f=-2U_0\frac{\sqrt3}{2} \] \[ U_f=-\sqrt3U_0 \] \(\textbf{Determine the work done by the field:}\) The reduced field remains unchanged during the stated rotation. Hence, \[ W_{\mathrm{field}}=U_i-U_f \] \[ W_{\mathrm{field}}=U_0-(-\sqrt3U_0) \] \[ W_{\mathrm{field}}=+(1+\sqrt3)U_0 \] \(\textbf{Physical interpretation:}\) The table determines the product \(pE\) without separately determining \(p\) and \(E\). Reducing the field changes both the torque scale and the energy scale by the same factor. The rotation moves the dipole towards a lower-energy, more aligned orientation, so the electric field does positive work. \(\textbf{Final answer:}\) \[ \boxed{\tau_{120}=\sqrt3U_0,\qquad W_{\mathrm{field}}=+(1+\sqrt3)U_0} \]

Explanation: \(\textbf{Determine the total reduction in separation:}\) The particles initially have separation \[ r_i=3a \] At the specified later instant, their separation is \[ r_f=a \] Since the charges are opposite, they attract each other and move towards one another. Therefore, the total reduction in their separation is \[ \Delta r=r_i-r_f \] \[ \Delta r=3a-a=2a \] Let \(s_1\) and \(s_2\) be the distances travelled by the lighter and heavier particles, respectively. Because the particles move towards each other, \[ s_1+s_2=2a \] \(\textbf{Apply conservation of centre-of-mass position:}\) Only mutual internal electrostatic forces act on the two-particle system. The total external force is zero. Since the system is initially at rest, its centre of mass remains stationary. Therefore, the centre-of-mass condition gives \[ ms_1=(3m)s_2 \] Cancelling \(m\), \[ s_1=3s_2 \] Thus, the lighter particle travels three times the distance travelled by the heavier particle. \(\textbf{Determine the individual displacements:}\) Substituting \(s_1=3s_2\) into the total-displacement relation, \[ 3s_2+s_2=2a \] \[ 4s_2=2a \] Therefore, \[ s_2=\frac a2 \] The lighter particle travels \[ s_1=3s_2 \] \[ s_1=\frac{3a}{2} \] \(\textbf{Calculate the initial electrostatic potential energy:}\) The interaction potential energy of two point charges is \[ U(r)=\frac{kq_1q_2}{r} \] The charges are \(+q\) and \(-2q\), so \[ U(r)=-\frac{2kq^2}{r} \] At the initial separation \(3a\), \[ U_i=-\frac{2kq^2}{3a} \] \(\textbf{Determine the potential energy at separation }a\textbf{:}\) At the later separation, \[ U_f=-\frac{2kq^2}{a} \] The change in potential energy is \[ \Delta U=U_f-U_i \] \[ \Delta U=-\frac{2kq^2}{a}+\frac{2kq^2}{3a} \] \[ \Delta U=-\frac{4kq^2}{3a} \] The negative change indicates that electrostatic potential energy decreases as the unlike charges approach each other. \(\textbf{Calculate the total work done by electrostatic forces:}\) The combined work done by the mutual electrostatic forces is \[ W_{\mathrm{total}}=-\Delta U \] Therefore, \[ W_{\mathrm{total}}=\frac{4kq^2}{3a} \] Both particles start from rest, so this work becomes their combined kinetic energy. \(\textbf{Determine how the work is distributed:}\) The system's total momentum remains zero because no external force acts on it. Therefore, the particles have equal and opposite momenta at every instant: \[ mv_1=(3m)v_2 \] Hence, \[ v_1=3v_2 \] By the work-energy theorem, the work done on each particle equals its gain in kinetic energy. For the lighter particle, \[ W_1=\frac12mv_1^2 \] For the heavier particle, \[ W_2=\frac12(3m)v_2^2 \] Their ratio is \[ \frac{W_1}{W_2}=\frac{mv_1^2}{3mv_2^2} \] Substituting \(v_1=3v_2\), \[ \frac{W_1}{W_2}=\frac{m(3v_2)^2}{3mv_2^2} \] \[ \frac{W_1}{W_2}=3 \] Thus, \[ W_1=3W_2 \] \(\textbf{Calculate the work done on the lighter particle:}\) The total electrostatic work is \[ W_{\mathrm{total}}=W_1+W_2 \] Using \(W_1=3W_2\), \[ W_{\mathrm{total}}=4W_2 \] Therefore, the lighter particle receives three-fourths of the total kinetic-energy gain: \[ W_1=\frac34W_{\mathrm{total}} \] Substituting the calculated total work, \[ W_1=\frac34\left(\frac{4kq^2}{3a}\right) \] Hence, \[ W_1=\frac{kq^2}{a} \] \(\textbf{Consistency check:}\) The heavier particle receives \[ W_2=\frac{W_1}{3}=\frac{kq^2}{3a} \] The combined work is \[ W_1+W_2=\frac{kq^2}{a}+\frac{kq^2}{3a} \] \[ W_1+W_2=\frac{4kq^2}{3a} \] This agrees with the decrease in electrostatic potential energy. The individual displacement ratio also agrees with the work ratio because the mutual electrostatic forces have equal magnitudes at every instant, and both particles move towards each other. \(\textbf{Physical interpretation:}\) The stationary centre of mass requires the lighter particle to travel farther than the heavier particle. As the unlike charges approach, their electrostatic potential energy decreases, and the electrostatic forces do positive work on both particles. The lighter particle receives the larger share of the released energy because it acquires greater speed under the equal-momentum condition. \(\textbf{Final answer:}\) \[ \boxed{s_1=\frac{3a}{2},\qquad W_1=\frac{kq^2}{a}} \]

Explanation: \(\textbf{Interpret the kinetic-energy decrease:}\) The particle moves outwards, but its kinetic energy decreases from \(3U_0\) to \(U_0\). Therefore, the electrostatic force does negative work during this displacement. A positive source attracts a negative particle, so the moving charge must be negative. \(\textbf{Relate the measurements through field work:}\) For a test charge \(q_t\), the work done by the point-source field between radial distances \(r_i\) and \(r_f\) is \[ W_{\mathrm{field}}=kQq_t\left(\frac1{r_i}-\frac1{r_f}\right) \] The work-energy theorem gives \[ K_f-K_i=W_{\mathrm{field}} \] Substituting the two measured energies and distances, \[ U_0-3U_0=kQq_t\left(\frac1a-\frac1{2a}\right) \] \[ -2U_0=\frac{kQq_t}{2a} \] Therefore, \[ q_t=-\frac{4aU_0}{kQ} \] \(\textbf{Determine the particle's total mechanical energy:}\) The interaction potential energy at radius \(r\) is \[ U(r)=\frac{kQq_t}{r} \] Using the inferred charge, \[ U(r)=-\frac{4aU_0}{r} \] At the initial measured position \(r=a\), \[ U(a)=-4U_0 \] The total mechanical energy is \[ E_{\mathrm{total}}=K(a)+U(a) \] \[ E_{\mathrm{total}}=3U_0-4U_0 \] \[ E_{\mathrm{total}}=-U_0 \] \(\textbf{Identify the outer turning point:}\) At the greatest separation, the radial speed becomes zero momentarily. Thus, \[ K(r_{\max})=0 \] Energy conservation requires \[ U(r_{\max})=E_{\mathrm{total}} \] \[ -\frac{4aU_0}{r_{\max}}=-U_0 \] Therefore, \[ r_{\max}=4a \] \(\textbf{Check the observed intermediate position:}\) At \(r=2a\), \[ U(2a)=-2U_0 \] Hence, \[ K(2a)=E_{\mathrm{total}}-U(2a) \] \[ K(2a)=-U_0+2U_0=U_0 \] This reproduces the second measurement. \(\textbf{Physical interpretation:}\) The measured kinetic-energy change identifies both the sign and magnitude of the unknown charge. The negative total mechanical energy means the particle cannot escape to infinity under the stated radial motion. Its outward movement terminates at the calculated turning point. \(\textbf{Final answer:}\) \[ \boxed{q_t=-\frac{4aU_0}{kQ},\qquad r_{\max}=4a} \]

Explanation: \(\textbf{Determine the mutual interaction energy:}\) The two source charges have opposite signs. Their separation is fixed by the rigid insulating rod: \[ r=2a \] Their mutual interaction potential energy is \[ U_{\mathrm{mutual}}=\frac{k(+2q)(-q)}{2a} \] \[ U_{\mathrm{mutual}}=-\frac{kq^2}{a} \] The rod is translated without changing its length. Therefore, the final mutual interaction energy is unchanged: \[ U_f=-\frac{kq^2}{a} \] \(\textbf{Distinguish mutual energy from external-field work:}\) Although the mutual interaction energy remains constant, both charges move through the applied uniform field. That field can perform work on the charges individually. The work contributions must be added algebraically. \(\textbf{Calculate the field work on }+2q\textbf{:}\) The uniform field and the displacement both point along positive \(x\). Thus, \[ W_+=q_+\vec E_u\cdot\Delta\vec r \] \[ W_+=(2q)\left(\frac{kq}{a^2}\hat i\right)\cdot(a\hat i) \] \[ W_+=+\frac{2kq^2}{a} \] \(\textbf{Calculate the field work on }-q\textbf{:}\) The negative charge experiences an electric force opposite to the field. Therefore, \[ W_-=(-q)\left(\frac{kq}{a^2}\hat i\right)\cdot(a\hat i) \] \[ W_-=-\frac{kq^2}{a} \] \(\textbf{Combine the external-field contributions:}\) \[ W_{\mathrm{field}}=W_++W_- \] \[ W_{\mathrm{field}}=\frac{2kq^2}{a}-\frac{kq^2}{a} \] \[ W_{\mathrm{field}}=+\frac{kq^2}{a} \] \(\textbf{Determine the quasistatic external work:}\) The rod undergoes no change in kinetic energy. The total work by the external agent therefore balances the field work: \[ W_{\mathrm{ext}}=-W_{\mathrm{field}} \] \[ W_{\mathrm{ext}}=-\frac{kq^2}{a} \] \(\textbf{Physical interpretation:}\) The internal Coulomb forces are equal and opposite, and the pair separation does not change. Hence, there is no change in mutual interaction potential energy. However, the charge pair has nonzero net charge, so a uniform external field exerts a nonzero resultant translational force. The agent does negative work while restraining the field-assisted translation. \(\textbf{Final answer:}\) \[ \boxed{U_f=-\frac{kq^2}{a},\qquad W_{\mathrm{ext}}=-\frac{kq^2}{a}} \]

Explanation: \(\textbf{Calculate the initial kinetic energy:}\) The lighter particle has mass \(m\) and speed \(2v\). The heavier particle has mass \(2m\) and speed \(v\). Their combined kinetic energy is \[ K_i=\frac12m(2v)^2+\frac12(2m)v^2 \] \[ K_i=2mv^2+mv^2 \] \[ K_i=3mv^2 \] Using the given speed condition, \[ K_i=\frac{3kq^2}{2a} \] \(\textbf{Calculate the initial interaction energy:}\) The particles have like charges. At separation \(4a\), \[ U_i=\frac{k(q)(2q)}{4a} \] \[ U_i=\frac{kq^2}{2a} \] \(\textbf{Determine total mechanical energy:}\) \[ E_{\mathrm{total}}=K_i+U_i \] \[ E_{\mathrm{total}}=\frac{3kq^2}{2a}+\frac{kq^2}{2a} \] \[ E_{\mathrm{total}}=\frac{2kq^2}{a} \] \(\textbf{Identify the minimum separation:}\) Because the initial total momentum is zero, the centre of mass is at rest. At the minimum separation, the particles are momentarily at rest relative to each other. Both speeds consequently vanish at that instant. All the mechanical energy is then interaction potential energy: \[ \frac{k(2q^2)}{r_{\min}}=\frac{2kq^2}{a} \] Therefore, \[ r_{\min}=a \] \(\textbf{Calculate the kinetic energy at separation }2a\textbf{:}\) At this intermediate separation, \[ U(2a)=\frac{2kq^2}{2a} \] \[ U(2a)=\frac{kq^2}{a} \] Hence, \[ K(2a)=E_{\mathrm{total}}-U(2a) \] \[ K(2a)=\frac{kq^2}{a} \] \(\textbf{Use momentum conservation to compare speeds:}\) Let the speed of the heavier particle at \(2a\) be \(u\). Since total momentum remains zero, \[ mu_1=(2m)u \] \[ u_1=2u \] The total kinetic energy at this separation is \[ K=\frac12m(2u)^2+\frac12(2m)u^2 \] \[ K=3mu^2 \] \(\textbf{Determine the heavier-particle speed:}\) \[ 3mu^2=\frac{kq^2}{a} \] Therefore, \[ u=\sqrt{\frac{kq^2}{3ma}} \] \(\textbf{Physical interpretation:}\) The particles initially approach despite their mutual repulsion. Their kinetic energy is converted into positive electrostatic interaction energy. Momentum conservation establishes their speed ratio throughout the motion and ensures that both momentarily stop at closest approach. \(\textbf{Final answer:}\) \[ \boxed{r_{\min}=a,\qquad u_{\mathrm{heavier}}(2a)=\sqrt{\frac{kq^2}{3ma}}} \]

Explanation: \(\textbf{Infer the horizontal field component:}\) The potential energy of an electric dipole in a uniform field is \[ U=-\vec p\cdot\vec E \] Write the unknown field as \[ \vec E=E_x\hat i+E_y\hat j \] For dipole moment \(p\hat i\), \[ -pE_x=-3U_0 \] Therefore, \[ E_x=\frac{3U_0}{p} \] \(\textbf{Infer the vertical field component:}\) For moment \(p\hat j\), \[ -pE_y=+4U_0 \] Hence, \[ E_y=-\frac{4U_0}{p} \] The reconstructed electric field is \[ \vec E=\frac{U_0}{p}(3\hat i-4\hat j) \] \(\textbf{Calculate its magnitude:}\) \[ E=\sqrt{E_x^2+E_y^2} \] \[ E=\frac{U_0}{p}\sqrt{3^2+(-4)^2} \] \[ E=\frac{5U_0}{p} \] \(\textbf{Evaluate the new dipole energy:}\) The final moment is \[ \vec p'=\frac p5(4\hat i+3\hat j) \] Thus, \[ U'=-\vec p'\cdot\vec E \] \[ U'=-\frac{U_0}{5}(4\hat i+3\hat j)\cdot(3\hat i-4\hat j) \] \[ U'=-\frac{U_0}{5}(12-12) \] \[ U'=0 \] The dipole moment is perpendicular to the inferred field. \(\textbf{Calculate the torque vector:}\) \[ \vec\tau'=\vec p'\times\vec E \] Substituting, \[ \vec\tau'=\frac{U_0}{5}(4\hat i+3\hat j)\times(3\hat i-4\hat j) \] Only the cross terms contribute: \[ \vec\tau'=\frac{U_0}{5}[(-16)\hat k-9\hat k] \] \[ \vec\tau'=-5U_0\hat k \] \(\textbf{Physical interpretation:}\) Two energy measurements along perpendicular directions determine the components of the unknown field. The final moment is perpendicular to that field, giving zero potential energy under the standard convention. Nevertheless, the torque magnitude is maximum and its negative \(z\)-direction follows from the vector cross product. \(\textbf{Final answer:}\) \[ \boxed{E=\frac{5U_0}{p},\qquad U'=0,\qquad \vec\tau'=-5U_0\hat k} \]

Explanation: \(\textbf{Write the two instrument equations:}\) The physical dipole energy is \[ U(\theta)=-pE\cos\theta \] The instrument adds a constant \(C\), so its readings need not equal the physical energies. At \(60^\circ\), \[ -\frac12pE+C=-U_0 \] At \(120^\circ\), \[ +\frac12pE+C=3U_0 \] \(\textbf{Separate the energy scale from the offset:}\) Subtracting the first equation from the second, \[ pE=3U_0-(-U_0) \] \[ pE=4U_0 \] Adding the two equations, \[ 2C=2U_0 \] Therefore, \[ C=U_0 \] \(\textbf{Determine the maximum torque magnitude:}\) The electric torque magnitude is \[ \tau=pE|\sin\theta| \] Its maximum occurs at perpendicular alignment: \[ |\sin\theta|=1 \] Thus, \[ \tau_{\max}=pE \] \[ \tau_{\max}=4U_0 \] \(\textbf{Determine the true initial energy:}\) At \(60^\circ\), \[ U_i=-pE\cos60^\circ \] \[ U_i=-\frac12(4U_0) \] \[ U_i=-2U_0 \] This equals the instrument reading minus its offset. \(\textbf{Determine the true final energy:}\) At \(120^\circ\), \[ U_f=-pE\cos120^\circ \] \[ U_f=+\frac12(4U_0) \] \[ U_f=+2U_0 \] \(\textbf{Calculate electrostatic field work:}\) \[ W_{\mathrm{field}}=U_i-U_f \] \[ W_{\mathrm{field}}=-2U_0-2U_0 \] \[ W_{\mathrm{field}}=-4U_0 \] \(\textbf{Physical interpretation:}\) The additive calibration offset changes reported energy values but not their difference. It therefore leaves the inferred dipole torque and rotational field work unchanged. The rotation raises the dipole's potential energy, so the electric field does negative work. \(\textbf{Final answer:}\) \[ \boxed{\tau_{\max}=4U_0,\qquad W_{\mathrm{field}}=-4U_0} \]

Explanation: The Assertion is correct because an electric dipole's equilibrium stability depends on its orientation relative to the electric field. Initially, the moment is parallel to the field, corresponding to minimum potential energy under \(U=-pE\cos\theta\). Reversing the external field without rotating the dipole changes the relative orientation to antiparallel. At exact antiparallel alignment, the angle is \(180^\circ\). The torque magnitude is \(pE|\sin180^\circ|=0\), so the dipole does not acquire an instantaneous turning tendency if it is exactly aligned and undisturbed. However, antiparallel alignment corresponds to maximum potential energy. If the dipole is displaced through a small angle, its potential energy decreases. The resulting electric torque tends to carry it farther from the antiparallel configuration, rather than restoring it. That behaviour defines an unstable equilibrium. The parallel state, by contrast, is stable because a small angular displacement raises the potential energy and produces a restoring torque. The Reason correctly identifies the relative-orientation change and the response to a small perturbation. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Compare the endpoint energies:}\) The dipole potential energy is \[ U=-pE\cos\theta \] Initially, \[ \theta_i=+60^\circ \] Therefore, \[ U_i=-pE\cos60^\circ \] \[ U_i=-\frac12pE \] Finally, \[ \theta_f=-60^\circ \] Because cosine is an even function, \[ \cos(-60^\circ)=\cos60^\circ \] Thus, \[ U_f=-\frac12pE \] \(\textbf{Determine the total field work:}\) \[ W_{\mathrm{field}}=U_i-U_f \] \[ W_{\mathrm{field}}=0 \] The equal endpoint energies make the net work zero. \(\textbf{Find the signed torque relation:}\) In the \(xy\)-plane, \[ \vec p=p(\cos\theta\hat i+\sin\theta\hat j) \] The applied field is \(E\hat i\). Therefore, \[ \vec\tau=\vec p\times\vec E \] \[ \vec\tau=-pE\sin\theta\,\hat k \] \(\textbf{Determine the initial torque:}\) At \(+60^\circ\), \[ \vec\tau_i=-pE\sin60^\circ\,\hat k \] \[ \vec\tau_i=-\frac{\sqrt3}{2}pE\hat k \] \(\textbf{Determine the final torque:}\) At \(-60^\circ\), \[ \sin(-60^\circ)=-\frac{\sqrt3}{2} \] Therefore, \[ \vec\tau_f=+\frac{\sqrt3}{2}pE\hat k \] \(\textbf{Physical interpretation:}\) The two orientations are equally inclined to the field on opposite sides of its direction. Their energies are equal, but their torques act in opposite rotational senses. Zero net work between the orientations does not imply zero torque throughout the intervening rotation. \(\textbf{Final answer:}\) \[ \boxed{W_{\mathrm{field}}=0,\qquad \vec\tau_i=-\frac{\sqrt3}{2}pE\hat k,\qquad \vec\tau_f=+\frac{\sqrt3}{2}pE\hat k} \]

Explanation: \(\textbf{Determine the initial potential energy:}\) The dipole potential energy is \[ U(\theta)=-pE\cos\theta \] At the initial angle \(120^\circ\), \[ U_i=-pE\cos120^\circ \] \[ U_i=+\frac12pE \] \(\textbf{Calculate total mechanical energy:}\) The initial rotational kinetic energy is \[ K_i=\frac14pE \] Therefore, \[ E_{\mathrm{total}}=K_i+U_i \] \[ E_{\mathrm{total}}=\frac14pE+\frac12pE \] \[ E_{\mathrm{total}}=\frac34pE \] \(\textbf{Check whether antiparallel alignment is reachable:}\) At \(180^\circ\), the potential energy is \[ U(180^\circ)=+pE \] Since \[ \frac34pE\lt pE \] the dipole does not possess enough mechanical energy to reach exact antiparallel alignment. It must reverse its rotation at a smaller angle. \(\textbf{Determine the turning angle:}\) At the turning point, the angular speed is zero. Thus, \[ K_{\mathrm{turn}}=0 \] Energy conservation requires \[ -pE\cos\theta_{\max}=\frac34pE \] Cancelling \(pE\), \[ \cos\theta_{\max}=-\frac34 \] Therefore, \[ \theta_{\max}=\cos^{-1}\left(-\frac34\right) \] This is larger than \(120^\circ\) but smaller than \(180^\circ\). \(\textbf{Calculate the rotational energy at }90^\circ\textbf{:}\) After reversing direction, the dipole can rotate towards smaller angles. At \(90^\circ\), \[ U_{90}=0 \] Therefore, \[ K_{90}=E_{\mathrm{total}}-U_{90} \] \[ K_{90}=\frac34pE \] \(\textbf{Determine the angular speed:}\) Rotational kinetic energy is \[ K_{90}=\frac12I\omega_{90}^2 \] Hence, \[ \frac12I\omega_{90}^2=\frac34pE \] \[ \omega_{90}^2=\frac{3pE}{2I} \] \[ \omega_{90}=\sqrt{\frac{3pE}{2I}} \] \(\textbf{Physical interpretation:}\) An initially increasing angle raises the dipole's potential energy and reduces its rotational kinetic energy. The available total energy determines the turning point. After reversal, decreasing potential energy is converted into rotational kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{\theta_{\max}=\cos^{-1}\left(-\frac34\right),\qquad \omega_{90}=\sqrt{\frac{3pE}{2I}}} \]