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)

Explanation:

Given that,

Electric field = 6500 N/C

Charge

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

Put the value into the formula

The change in electric potential energy is .

(b). When a charge is moved in the negative x direction

We need to calculate the electric potential energy

Using formula of potential energy

Put the value into the formula

The change in electric potential energy is .

(c). When a charge is moved in the positive y direction

We need to calculate the electric potential energy

Using formula of potential energy

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.

graphing motion: mastery test

209be

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0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2