Question

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy of the pair of charges is 5.4 x 10^-8J. When the second charge is moved to point b, the electric force on the charge does -1.9x10^-8J of work. What is the electric potential energy of the pair of charges when the second charge is at point b? Please explain the answer briefly

Answer 1

`U_(b)=+7.3*10^(-8)J`

Step-by-step explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it
`W_(a-b)=-1.9*10^(-8)J`

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

`W_(a-b)=U_(a)-U_(b)\\U_(b)=U_(a)-W_(a-b)`

Now substitute the given values

So

`U_(b)=5.4*10^(-8)J-(-1.9*10^(-8)J)\\U_(b)=+7.3*10^(-8)J`