Question

A uniform electric field with a magnitude of 6500 N/C points in the positive x direction. a) Find the change in electric potential energy when a +12.5-μC charge is moved 6.00 cm in the positive x direction. (J)

b) Find the change in electric potential energy when a +12.5-μC charge is moved 6.00 cm in the negative x direction. (J)
c) Find the change in electric potential energy when a +12.5-μC charge is moved 6.00 cm in the positive y direction. (J)

Answer 1

Step-by-step explanation:

Given that,

Electric field = 6500 N/C

Charge
`q=+12.5\ \mu C`

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`

`\Delta U=-F\cdot d`

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

Put the value into the formula

`\Delta U=-12.5*10^(-6)*6500*6.00*10^(-2)`

`\Delta U=-4.88*10^(-3)\ J`

The change in electric potential energy is
`-4.88*10^(-3)\ J`.

(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)`

Put the value into the formula

`\Delta U=12.5*10^(-6)*6500*6.00*10^(-2)`

`\Delta U=4.88*10^(-3)\ J`

The change in electric potential energy is
`4.88*10^(-3)\ J`.

(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)`

`\Delta U=0`

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.