Question

Two small, identical metal balls with charges 7.0 µC and 14.0 µC are held in place 1.9 m apart. In an experiment, they are connected for a short time by a conducting wire.
(a) What will be the charge on each ball after this experiment?

1. new charge on 7.0 µC ball = ? µC

2. ew charge on 14.0 µC ball = ? µC


(b) By what factor will the magnitude of the electrostatic force on either ball change after this experiment is performed?


= ?

Answer 1

The new charge on 7.0 µC ball is 10.5 µC and the new charge on 14.0 µC ball is 10.5 µC.

The change in the electrostatic force after the experiment is 0.031 N.

The given parameters:

• Charge on first metal ball, q1 = 7.0 µC
• Charge on second metal ball, q2 = 14 µC
• Distance between the charges, r = 1.9 m

Charge on each ball after experiment

After the experiment the charges will be at equilibrium, and the charge on each metal ball will be equal.

$Q_t = q_1 + q_2\\\\Q_t = 7 \mu C + 14 \mu C\\\\Q_t = 21 \ \mu C$

$Q_1 = Q_2 = \frac{Q_t}{2} = \frac{21 \mu C}{2} = 10.5 \ \mu C$

The new charge on 7.0 µC ball = 10.5 µC

The new charge on 14.0 µC ball = 10.5 µC

Change in electrostatic force

$F_1 = \frac{kQ_1Q_2}{r^2} \\\\F_1 = \frac{9\times 10^9}{1.9^2} (7 \times 10^{-6} \times 14 \times 10^{-6}) \ \\\\ F_1 = 0.244 \ N\\\\for \ new \ charges;\\\\F_2 = \frac{kQ_1Q_2}{r^2} \\\\F_2 = \frac{9\times 10^9}{1.9^2} (10.5 \times 10^{-6} \times 10.5 \times 10^{-6}) \ \\\\ F_2 = 0.275 \ N\\\\\Delta F = F_2 - F_2\\\\\Delta F = 0.275 \ N \ - \ 0.244 \ N\\\\\Delta F = 0.031 \ N$

Lear more about electrostatic force here: brainly.com/question/17692887