Explanation:
Given that,
Electric field = 6500 N/C
Charge ![q=+12.5\ \mu C](https://tex.z-dn.net/?f=q%3D%2B12.5%5C%20%5Cmu%20C)
Distance = 6.00 cm
(a). When a charge is moved in the positive x direction
We need to calculate the electric potential energy
Using formula of potential energy
![\Delta U=-W](https://tex.z-dn.net/?f=%5CDelta%20U%3D-W)
![\Delta U=-F\cdot d](https://tex.z-dn.net/?f=%5CDelta%20U%3D-F%5Ccdot%20d)
![\Delta U=q(E\cdot d)](https://tex.z-dn.net/?f=%5CDelta%20U%3Dq%28E%5Ccdot%20d%29)
Put the value into the formula
![\Delta U=-12.5\times10^{-6}\times6500\times6.00\times10^{-2}](https://tex.z-dn.net/?f=%5CDelta%20U%3D-12.5%5Ctimes10%5E%7B-6%7D%5Ctimes6500%5Ctimes6.00%5Ctimes10%5E%7B-2%7D)
![\Delta U=-4.88\times10^{-3}\ J](https://tex.z-dn.net/?f=%5CDelta%20U%3D-4.88%5Ctimes10%5E%7B-3%7D%5C%20J)
The change in electric potential energy is
.
(b). When a charge is moved in the negative x direction
We need to calculate the electric potential energy
Using formula of potential energy
![\Delta U=q(E\cdot d)](https://tex.z-dn.net/?f=%5CDelta%20U%3Dq%28E%5Ccdot%20d%29)
Put the value into the formula
![\Delta U=12.5\times10^{-6}\times6500\times6.00\times10^{-2}](https://tex.z-dn.net/?f=%5CDelta%20U%3D12.5%5Ctimes10%5E%7B-6%7D%5Ctimes6500%5Ctimes6.00%5Ctimes10%5E%7B-2%7D)
![\Delta U=4.88\times10^{-3}\ J](https://tex.z-dn.net/?f=%5CDelta%20U%3D4.88%5Ctimes10%5E%7B-3%7D%5C%20J)
The change in electric potential energy is
.
(c). When a charge is moved in the positive y direction
We need to calculate the electric potential energy
Using formula of potential energy
![\Delta U=q(E\cdot d)](https://tex.z-dn.net/?f=%5CDelta%20U%3Dq%28E%5Ccdot%20d%29)
![\Delta U=0](https://tex.z-dn.net/?f=%5CDelta%20U%3D0)
Because the electric field direction is perpendicular to the movement.
So, The change in electric potential energy is zero.
Hence, This is the required solution.