Question

In case A two negative charges which are equal in magnitude are separated by a distance d. In case B the same charges are separated by a distance 2d. Which configuration has the highest potential energy

Answer 1

Case A

Step-by-step explanation:

We are given that

Case A:

Let

`q_1=q_2=-q`

Distance between two negative charges,r=d

Case B:

Charges are same

Distance between two charges,r=2d

Potential energy=
`U=(kq_1q_2)/(r)`

Using the formula

For case A

`U_A=(kq^2)/(d)`

For case B

`U_B=(Kq^2)/(2d)`

Potential energy is inversely proportional to distance between two charges

The distance between two charges in case A is small therefore, it has the highest potential energy.