Question

How much work is done in moving a charge of $3C $ from a point at the volts $115 $ to a point at $125V $?

Answer 1

Hint The given question is an easy question. It is a direct question where the value of the charge and the different potential of the charge at two points is provided. We have to determine the work done in moving the charge from one point to another. Find the change in potential and use the work done formula directly to get the answer.

Complete step by step answer
Potential energy per unit charge is called an electric potential.
The amount of work done in bringing the point charge from infinity to the point is called the electric potential at a point. Work is done against the electric field.
The amount of work done in bringing the point charge from one point to another point is called the electric potential difference between two points. Work is done against the electric field.
 $ \Rightarrow V = Wq $
Where,
V is the potential
W is the work done
q is the charge
So, work done in moving a charge from one point to another point is determined by the product of charge and the change in potential from one point to another point
 $ \Rightarrow W = q\Delta V $
Where,
W is the work done
q is the charge
 $\Delta V $ is the change in the potential
Given,
The value of the given charge, $q = 3C $
The potential of the point where the charge is present at first, ${V_1} = 115V $
The potential of the point where the charge is moved, ${V_2} = 125V $
The change in the potential, $\Delta V = {V_2} - {V_1} $
 $ \Rightarrow \Delta V = {V_2} - {V_1} $
 $ \Rightarrow \Delta V = 125 - 115 $
 $ \Rightarrow \Delta V = 10V $
Now, we have seen above that work is done in moving the charge from one potential to another potential is given by
 $ \Rightarrow W = q\Delta V $
 $ \Rightarrow W = 3 \times 10 $
 $ \Rightarrow W = 30J $
Note The work done in bringing a charge from infinity will be zero since the charge is at infinity. The unit of work done is joule since work done is the amount of energy and energy is expressed in terms of joule