Physics, 22.08.2019 16:30 nikki225

Auniform 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)

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Answer from: tiff3692

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

$\Delta U=-12.5\times10^{-6}\times6500\times6.00\times10^{-2}$

$\Delta U=-4.88\times10^{-3}\ J$

The change in electric potential energy is $-4.88\times10^{-3}\ J$ .

(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

$\Delta U=12.5\times10^{-6}\times6500\times6.00\times10^{-2}$

$\Delta U=4.88\times10^{-3}\ J$

The change in electric potential energy is $4.88\times10^{-3}\ J$ .

(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.

Answer from: Quest

answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.

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Auniform electric field with a magnitude of 6500 n/c points in the positive x direction. a) find the...

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