Question

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential energy when a +10.5-μC charge is moved 5.50 cm in BOTH the negative and positive x direction.

Answer 1

Step-by-step explanation:

Given that,

Electric field = 5750 N/C

Charge
`q=+10.5*10^(-6)\ C`

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

`\Delta U=-W`

`\Delta U=-F\cdot d`

`\Delta U=-q(E\cdot d)`

Put the value into the formula

`\Delta U=-10.5*10^(-6)*5750*5.50*10^(-2)`

`\Delta U=-3.32*10^(-3)\ J`

The change in electric potential energy is
`-3.32*10^(-3)\ J`

(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

`\Delta U=-W`

`\Delta U=-F\cdot (-d)`

`\Delta U=-q(E\cdot (-d))`

Put the value into the formula

`\Delta U=-10.5*10^(-6)*5750*(-5.50*10^(-2))`

`\Delta U=3.32*10^(-3)\ J`

The change in electric potential energy is
`3.32*10^(-3)\ J`

Hence, This is the required solution.