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atroni [7]
3 years ago
9

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.
Physics
1 answer:
castortr0y [4]3 years ago
3 0

Explanation:

Given that,

Electric field = 5750 N/C

Charge q=+10.5\times10^{-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\times10^{-6}\times5750\times5.50\times10^{-2}

\Delta U=-3.32\times10^{-3}\ J

The change in electric potential energy  is -3.32\times10^{-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\times10^{-6}\times5750\times(-5.50\times10^{-2})

\Delta U=3.32\times10^{-3}\ J

The change in electric potential energy  is 3.32\times10^{-3}\ J

Hence, This is the required solution.

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