Question

Two small identical conducting spheres are placed with their centers 0.41 m apart. One is given a charge of 12 ✕ 10−9 C, the other a charge of −23 ✕ 10−9 C. (a) Find the electrostatic force exerted on one sphere by the other. magnitude 1.48e-5 Correct: Your answer is correct. N direction attractive Correct: Your answer is correct. (b) The spheres are connected by a conducting wire. Find the electrostatic force between the two after equilibrium is reached, where both spheres have the same charge.

Answer 1

(a) $-1.48\cdot 10^{-5}N$

The electrostatic force exerted between the two sphere is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges on the two spheres

r is the separation between the centres of the two spheres

In this problem,

$q_1 = 12\cdot 10^{-9} C\\q_2 = -23\cdot 10^{-9} C\\r = 0.41 m$

Substituting these values into the equation, we find the force

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(12\cdot 10^{-9}C)(-23\cdot 10^{-9} C)}{(0.41 m)^2}=-1.48\cdot 10^{-5}N$

And the negative sign means the force is attractive, since the two spheres have charges of opposite sign.

(b) $+1.62\cdot 10^{-6}N$

The total net charge over the two sphere is:

$Q=q_1 +q_2 = 12\cdot 10^{-9}C+(-23\cdot 10^{-9}C)=-11\cdot 10^{-9} C$

When the two spheres are connected, the charge distribute equally over the two spheres (since they are identical, they have same capacitance), so each sphere will have a charge of

$q=\frac{Q}{2}=\frac{-11\cdot 10^{-9}C}{2}=-5.5\cdot 10^{-9}C$

So the electrostatic force between the two spheres will now be

$F=k\frac{q^2}{r^2}$

And substituting numbers, we find

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(-5.5\cdot 10^{-9} C)^2}{(0.41 m)^2}=+1.62\cdot 10^{-6}N$

and the positive sign means the force is repulsive, since the two spheres have same sign charges.