Question

Two identical metal balls of radii 2.50

Answer 1

Answer:

+1.33 × $10^{-7}$ C and -1.33 × $10^{-7}$ C respectively.

Explanation:

Electric potential (V) is the work done in moving a unit positive charge from infinity to a reference point within an electric field. It is measured in volts.

     V = $\frac{kq}{r}$ ............. 1

where: k is a constant = 9 × $10^{9}$ N$m^{2} C^{-2}$, q is the charge and r is the distance between the charges.

From equation 1,

   q = $\frac{Vr}{k}$ ............... 2

The charge on each ball can be determined as;

given that; V = 1.2 × $10^{3}$, k = 9 × $10^{9}$ N$m^{2} C^{-2}$ and r = 1.00 m.

From equation 2,

  q = $\frac{1.2*10^{3} * 1.0}{9*10^{9} }$

     = 1.33 × $10^{-7}$ C

Thus, the charge on the first ball is +1.33 × $10^{-7}$ C, while the charge on the second ball is -1.33 × $10^{-7}$ C.