Question

Two identical conducting spheres carry charges of +5.0 μC and –1.0 μC, respectively. The centers of the spheres are initially separated by a distance L. The two spheres are brought together so that they are in contact. The spheres are then returned to their original separation L. What is the ratio of the magnitude of the electric force on either sphere after the spheres are touched to that before they were touched? A) 1/1 B) 4/5 C) 9/5 D) 5/1 E) 4/9

Answer 1

Answer:

B) 4/5

Explanation:

The magnitude of the electric force between the two spheres is given by

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

where

k is the Coulombs' constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

Initially, we have

$q_1 = 5.0\mu C=5.0\cdot 10^{-6}C\\q_2 = 1.0 \mu C=1.0 \cdot 10^{-6}C\\r = L$

So the force is

$F_1=k\frac{(5.0\cdot 10^{-6}C)(1.0\cdot 10^{-6}C)}{L^2}=(5.0 \cdot 10^{-12} C^2)\frac{k}{L^2}$

Later, the two spheres are brought together so they are in contact: this means that the total charge will redistribute equally on the two spheres (because they are identical).

The total charge is

$Q=q_1 + q_2 = +4.0 \mu C=4.0\cdot 10^{-6}C$

So each sphere will have a charge of

$q=\frac{Q}{2}=2.0\cdot 10^{-6} C$

So, the new force will be

$F_2=k\frac{(2.0\cdot 10^{-6}C)(120\cdot 10^{-6}C)}{L^2}=(4.0 \cdot 10^{-12} C^2)\frac{k}{L^2}$

And so the ratio of the two forces is

$\frac{F_2}{F_1}=\frac{4.0\cdot 10^{-12} C}{5.0\cdot 10^{-12} C}=\frac{4}{5}$