Question

How much work does the charge escalator do to move 2.30 μC of charge from the negative terminal to the positive terminal of a 3.00 V battery?

Answer 1

Answer:

$6.9\times 10^{-6} J$

Explanation:

We are given that

Charge=q=$2.3\mu C=2.3\times 10^{-6} C$

$1\mu C=10^{-6} C$

Potential difference=V=3 V

We know that

Work done=$V\times q$

Using the formula

Work done by charge to move from the negative terminal to the positive terminal of battery=$2.3\times 10^{-6}\times 3$ J

Work done by charge to move from the negative terminal to the positive terminal of battery=$6.9\times 10^{-6} J$