SATHEE: Electrostatics Question 221

Electrostatics Question 221

Question: Two equally charged, identical metal spheres A and B repel each other with a force ‘F’. The spheres are kept fixed with a distance ‘r’ between them. A third identical, but uncharged sphere C is brought in contact with A and then placed at the mid-point of the line joining A and B. The magnitude of the net electric force on C is [UPSEAT 2004; DCE 2005]

Options:

A) F

B) 3F/4

C) F/2

D) F/4

Show Answer

Answer:

Correct Answer: A

Solution:

Initially $F = k \frac{Q^{2}}{r^{2}}$ ……. (i)

Finally Force on C due to A, $F_{A} = \frac{k {(Q / 2)}^{2}}{{(r / 2)}^{2}} = \frac{k Q^{2}}{r^{2}}$

Force on C due to B, $F_{B} = \frac{K Q (Q / 2)}{{(r / 2)}^{2}} = \frac{2 K Q^{2}}{r^{2}}$

Net force on C, $F_{n e t} = F_{B} - F_{A} = \frac{k Q^{2}}{r^{2}} = F$

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Electrostatics Question 221

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