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Electrostatics question

2025 · Shift 2 · Q37
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Electrostatics question

2025 · Shift 2 · Q37

JEE AdvancedPhysicsElectrostaticsMultiple correct+4 / −2
A positive point charge of 10−810^{-8}10−8 C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking 14πϵ0=9×109\frac{1}{4\pi\epsilon_0} = 9 \times 10^94πϵ0​1​=9×109 Nm 2^22/C 2^22(where ϵ0\epsilon_0ϵ0​ is the permittivity of free space), which of the following statements is/are correct:
  1. A
    Before the grounding, the electrostatic potential of the sphere is 450 V.
  2. B
    Charge flowing from the sphere to the ground because of grounding is 5×10−95 \times 10^{-9}5×10−9 C.
  3. C
    After the grounding is removed, the charge on the sphere is −5×10−9-5 \times 10^{-9}−5×10−9 C.
  4. D
    The final electrostatic potential of the sphere is 300 V.
View written solutionFree

Correct answer: A, B, C

The problem asks us to evaluate four statements about a conducting sphere near a point charge, undergoing a sequence of operations: grounding, un-grounding, and moving the charge.

Given values:

  • Point charge, q=10−8q = 10^{-8}q=10−8 C
  • Initial distance of charge from sphere's center, d=20d = 20d=20 cm = 0.2 m
  • Radius of the sphere, R=10R = 10R=10 cm = 0.1 m
  • Coulomb's constant, k=14πϵ0=9×109k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9k=4πϵ0​1​=9×109 Nm2^22/C2^22

Step 1: Analyze the initial situation (before grounding) - Option A

  1. Initially, the conducting sphere is neutral. The external point charge qqq induces charges on the surface of the sphere, but the net charge on the sphere is zero.
  2. A conducting sphere is an equipotential body. The potential is the same at every point on its surface and inside it. We can conveniently calculate the potential at the center of the sphere.
  3. The potential at the center (VcenterV_{center}Vcenter​) is the sum of the potential due to the external point charge qqq (VqV_qVq​) and the potential due to the induced charges on the sphere's surface (VindV_{ind}Vind​).
  4. The potential at the center due to the external charge qqq at distance ddd is: Vq=kqdV_q = \frac{kq}{d}Vq​=dkq​
  5. The potential at the center due to the induced charges on the surface (each at a distance RRR from the center) is: Vind=kQindRV_{ind} = \frac{k Q_{ind}}{R}Vind​=RkQind​​ Since the sphere is initially neutral, the total induced charge Qind=0Q_{ind} = 0Qind​=0. Therefore, Vind=0V_{ind} = 0Vind​=0.
  6. The potential of the sphere is equal to the potential at its center: Vsphere=Vcenter=Vq+Vind=kqd+0V_{sphere} = V_{center} = V_q + V_{ind} = \frac{kq}{d} + 0Vsphere​=Vcenter​=Vq​+Vind​=dkq​+0
  7. Substituting the values: Vsphere=(9×109 Nm2/C2)×(10−8 C)0.2 m=900.2 V=450 VV_{sphere} = \frac{(9 \times 10^9 \text{ Nm}^2/\text{C}^2) \times (10^{-8} \text{ C})}{0.2 \text{ m}} = \frac{90}{0.2} \text{ V} = 450 \text{ V}Vsphere​=0.2 m(9×109 Nm2/C2)×(10−8 C)​=0.290​ V=450 V

Conclusion for A: Statement A is correct.

Step 2: Analyze the situation during grounding - Options B and C

  1. When the sphere is grounded, its potential becomes zero (Vsphere′=0V'_{sphere} = 0Vsphere′​=0). To achieve this, charge flows between the sphere and the ground.
  2. Let the charge on the sphere after grounding be q′q'q′. The potential of the sphere is now the sum of the potential due to the external charge qqq and the potential due to its own charge q′q'q′.
  3. Again, calculating the potential at the center: Vsphere′=Vcenter=Vdue to q+Vdue to q′=0V'_{sphere} = V_{center} = V_{\text{due to } q} + V_{\text{due to } q'} = 0Vsphere′​=Vcenter​=Vdue to q​+Vdue to q′​=0 Vsphere′=kqd+kq′R=0V'_{sphere} = \frac{kq}{d} + \frac{kq'}{R} = 0Vsphere′​=dkq​+Rkq′​=0
  4. We can solve for q′q'q′: kq′R=−kqd  ⟹  q′=−qRd\frac{kq'}{R} = -\frac{kq}{d} \implies q' = -q \frac{R}{d}Rkq′​=−dkq​⟹q′=−qdR​
  5. Substituting the values: q′=−(10−8 C)×0.1 m0.2 m=−0.5×10−8 C=−5×10−9 Cq' = -(10^{-8} \text{ C}) \times \frac{0.1 \text{ m}}{0.2 \text{ m}} = -0.5 \times 10^{-8} \text{ C} = -5 \times 10^{-9} \text{ C}q′=−(10−8 C)×0.2 m0.1 m​=−0.5×10−8 C=−5×10−9 C
  6. Evaluate Option C: When the grounding is removed, the sphere becomes isolated again. The charge q′q'q′ is trapped on it. So, the charge on the sphere after the grounding is removed is −5×10−9-5 \times 10^{-9}−5×10−9 C. Conclusion for C: Statement C is correct.
  7. Evaluate Option B: The initial charge on the sphere was 0. The final charge is q′=−5×10−9q' = -5 \times 10^{-9}q′=−5×10−9 C. This charge came from the ground. So, the charge that flowed to the sphere from the ground is −5×10−9-5 \times 10^{-9}−5×10−9 C. The question asks for the charge flowing from the sphere to the ground. This is the negative of the charge that flowed from the ground to the sphere. Charge from sphere to ground = −(charge from ground to sphere)=−(q′−0)=−(−5×10−9 C)=5×10−9 C-(\text{charge from ground to sphere}) = - (q' - 0) = -(-5 \times 10^{-9} \text{ C}) = 5 \times 10^{-9} \text{ C}−(charge from ground to sphere)=−(q′−0)=−(−5×10−9 C)=5×10−9 C. Conclusion for B: Statement B is correct.

Step 3: Analyze the final situation - Option D

  1. After the grounding is removed, the sphere has a fixed charge q′=−5×10−9q' = -5 \times 10^{-9}q′=−5×10−9 C.
  2. The point charge qqq is then moved to a new distance d′=20 cm+10 cm=30 cm=0.3d' = 20 \text{ cm} + 10 \text{ cm} = 30 \text{ cm} = 0.3d′=20 cm+10 cm=30 cm=0.3 m from the center of the sphere.
  3. The final potential of the sphere (VfinalV_{final}Vfinal​) is the sum of the potential due to its own charge q′q'q′ and the potential due to the external charge qqq at its new position d′d'd′.
  4. Calculating the potential at the center: Vfinal=Vdue to q′+Vdue to q at d′=kq′R+kqd′V_{final} = V_{\text{due to } q'} + V_{\text{due to } q \text{ at } d'} = \frac{kq'}{R} + \frac{kq}{d'}Vfinal​=Vdue to q′​+Vdue to q at d′​=Rkq′​+d′kq​
  5. Substituting the values: Vfinal=(9×109)(−5×10−9)0.1+(9×109)(10−8)0.3V_{final} = \frac{(9 \times 10^9)(-5 \times 10^{-9})}{0.1} + \frac{(9 \times 10^9)(10^{-8})}{0.3}Vfinal​=0.1(9×109)(−5×10−9)​+0.3(9×109)(10−8)​ Vfinal=−450.1+900.3=−450 V+300 V=−150 VV_{final} = \frac{-45}{0.1} + \frac{90}{0.3} = -450 \text{ V} + 300 \text{ V} = -150 \text{ V}Vfinal​=0.1−45​+0.390​=−450 V+300 V=−150 V

Conclusion for D: The final electrostatic potential of the sphere is -150 V. Statement D, which claims it is 300 V, is incorrect.

Final Summary

  • Statement A is correct.
  • Statement B is correct.
  • Statement C is correct.
  • Statement D is incorrect.
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