A positive charge of 6 C is placed in an electr... - SS2 Physics Electricity and Magnetism Question
A positive charge of +6 μC is placed in an electric field where the electric potential is 120 V. What is the change in electric potential energy if the charge is moved 4 cm in the direction opposite to the electric field?
The change in electric potential energy can be calculated using the formula:
ΔPE = q x ΔV, where ΔPE is the change in potential energy, q is the charge, and ΔV is the change in potential.
Plugging in the values, we have:
ΔPE = 6 x 10-6 C x (-120 V)
ΔPE = -7.2 x 10-4 J
Please share this, thanks:
Add your answer
No responses