Explanation:
Given that,
Electric field = 5750 N/C
Charge [tex]q=+10.5\times10^{-6}\ C[/tex]
Distance = 5.50 cm
(a). When the charge is moved in the positive x- direction
We need to calculate the change in electric potential energy
Using formula of electric 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=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}[/tex]
[tex]\Delta U=-3.32\times10^{-3}\ J[/tex]
The change in electric potential energy is [tex]-3.32\times10^{-3}\ J[/tex]
(b). When the charge is moved in the negative x- direction
We need to calculate the change in electric potential energy
Using formula of electric 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=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})[/tex]
[tex]\Delta U=3.32\times10^{-3}\ J[/tex]
The change in electric potential energy is [tex]3.32\times10^{-3}\ J[/tex]
Hence, This is the required solution.
