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

Respuesta :

Explanation:

It is given that,

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

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

[tex]W=(V_B-V_A)q\\\\W=-1000\times 10^{-6}\\\\W=E=-10^{-3}\ J[/tex]

So, the energy of charge decreases by [tex]10^{-3}\ J[/tex]. Hence, the correct option  is (a).