Question

Two small, identical conducting spheres repel each other with a force of 0.035 N when they are 0.35 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.055 N. What is the original charge on each sphere? (Enter the magnitudes in C.)

Answer 1

Explanation:

Given:

distance between two sphere =0.35 m

Electrical repel force =0.035 N

Electrical repel force after connecting wire =0.055 N.

The electrical force between the two spheres:

$F=k \frac{q_{1} q_{2}}{r^{2}}$

The electrical force between the two spheres after the wire is attached and removed:

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

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

$q=r \sqrt{\frac{F}{k}}=0.35 \times \sqrt{\frac{0.055}{8.99 \times 10^{9}}}=6.46 \times 10^{-7} \mathrm{C}$

So the total charge of the two spheres $=2 q=2 \times 6.46 \times 10^{-7}=1.29 \times 10^{-6} \mathrm{C}$

Then before connecting the wire, one sphere charge was $q[tex] and the charge of the other sphere was [tex]\left(1.29 \times 10^{-6}-q\right)$

The electrical force between the two spheres before connecting the wire:

$F=k \frac{q\left(1.29 \times 10^{-6}-q\right)}{r^{2}}$

$q\left(1.29 \times 10^{-6}-q\right)=\frac{F r^{2}}{k}=\frac{0.035 \times(0.35)^{2}}{8.99 \times 10^{9}}=0.348 \times 10^{-12}$

$-q^{2}+\left(1.29 \times 10^{-6}\right) q-\left(0.348 \times 10^{-12}\right)=0$

$-q^{2}+\left(1.29 \times 10^{-6}\right) q-\left(0.348 \times 10^{-12}\right)=0$