Question

A positive charge of 1 µC is taken from points A & B such that VA-VB = 100 V. Then the

energy of charge increases by 10-3 J
energy of charge decreases by 10-3 J
energy of charge remains unchanged
energy of charge decreases by 103

Answer 1

Step-by-step explanation:

It is given that,

Let Charge of
`1\ \mu C` is taken from points A & B such that
`V_A-V_B=1000\ V`.

We need to find the energy of charge. Electric potential is defined as the work done per unit of electric charge. So,

`W=(V_B-V_A)q\\\\W=-1000* 10^(-6)\\\\W=E=-10^(-3)\ J`

So, the energy of charge decreases by
`10^(-3)\ J`. Hence, the correct option is (a).