Question

Two identically charged metal spheres A and B experience a force of $2\times {10}^{-5}$ Newton of repulsive nature. Another, similar uncharged metal spheres C is brought near A and after contact with A it is separated and now placed midway between A and B. The total force on this new spheres C in Newton is:

• A. $1\times {10}^{-5}$
• B. $2\times {10}^{-5}$
• C. $0.5\times {10}^{-5}$
• D. $4\times {10}^{-5}$

Answer 1

Answer: (B)

$2\times {10}^{-5}$

Let the charges on the sphere A and B be Q,

Now C touches A and gets $Q/2$ charge so charge on A $=Q/2$

Now force on the A due to B $=\frac{K{Q}^2}{r^2}=2\times {10}^{-5}N$

now force on C due to A $=\frac{K(Q/2{)}^2}{\left(r/2\right)}$

now force on C due to B $=\frac{K\left(Q/2\right)\left(Q\right)}{(r/2{)}^2}$

net force $=\frac{K\left(Q/2\right)\left(Q\right)}{(r/2{)}^2}-$ $\frac{K(Q/2{)}^2}{\left(r/2\right)}$ $=\frac{K{Q}^2}{r^2}=2\times {10}^{-5}N$