Respuesta :
Explanation:
Given that,
Electric field = 6500 N/C
Charge [tex]q=+12.5\ \mu C[/tex]
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
[tex]\Delta U=-W[/tex]
[tex]\Delta U=-F\cdot d[/tex]
[tex]\Delta U=q(E\cdot d)[/tex]
Put the value into the formula
[tex]\Delta U=-12.5\times10^{-6}\times6500\times6.00\times10^{-2}[/tex]
[tex]\Delta U=-4.88\times10^{-3}\ J[/tex]
The change in electric potential energy is [tex]-4.88\times10^{-3}\ J[/tex].
(b). When a charge is moved in the negative x direction
We need to calculate the electric potential energy
Using formula of potential energy
[tex]\Delta U=q(E\cdot d)[/tex]
Put the value into the formula
[tex]\Delta U=12.5\times10^{-6}\times6500\times6.00\times10^{-2}[/tex]
[tex]\Delta U=4.88\times10^{-3}\ J[/tex]
The change in electric potential energy is [tex]4.88\times10^{-3}\ J[/tex].
(c). When a charge is moved in the positive y direction
We need to calculate the electric potential energy
Using formula of potential energy
[tex]\Delta U=q(E\cdot d)[/tex]
[tex]\Delta U=0[/tex]
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.
