A uniform electric field with a magnitude of 6100 V m−1 points in the positive x direction.

Find the change in electric potential energy when a +14 μC
charge is moved 5 cm in the positive x direction.

Answer in joules.

PE= -Eqd=61OO V/m * 12E-6C*.05, so the energy is reduced.

Electric field potential is defined as the negative of the work done by the electric field moving a charge in an electric field.

To find the change in electric potential energy, we can use the formula:

ΔPE = qΔV

where ΔPE is the change in electric potential energy, q is the charge, and ΔV is the change in electric potential.

In this case, the charge is +14 μC and it is moved 5 cm in the positive x direction. The uniform electric field is given as 6100 V/m, which represents the change in electric potential per meter.

First, we need to convert the charge from microcoulombs to coulombs:
14 μC = 14 × 10^-6 C

Next, we need to convert the distance moved from centimeters to meters:
5 cm = 5 × 10^-2 m

Now, we can calculate the change in electric potential:
ΔV = (electric field) × (distance)
ΔV = (6100 V/m) × (5 × 10^-2 m)

Multiply the values:
ΔV = 305 V

Finally, we can calculate the change in electric potential energy:
ΔPE = (charge) × (change in electric potential)
ΔPE = (14 × 10^-6 C) × (305 V)

Multiply the values:
ΔPE = 4.27 × 10^-3 J

Therefore, the change in electric potential energy is approximately 4.27 × 10^-3 joules (J).