Question

A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge
A. remains constant because the electric field is uniform.
B. increases because charge moves along the electric field.
C. decreases because charge moves along the electric field.
D. decreases because charge moves opposite to the electric field.

Answer 1

Hint:Use the concept of electric field, electrostatic force on a positive charge to determine the direction of motion of the positive charge. Also use the concept of conservation of energy in order to determine whether the potential energy of the positive charge increases, decreases or remains constant.

Complete answer:
When the positive charge is at rest, it has maximum potential energy. But when it is released from rest in the uniform electric field, it starts moving and its potential energy gets converted into kinetic energy. Therefore, the potential energy of the charge does not remain constant or increase. Hence, the options A and B are incorrect.
When the positive charge is released from rest in the uniform electric field, it moves along the direction of the electric field due to the electrostatic force acting on it and not in the opposite direction of the electric field.Hence, the option D is incorrect.
The positive charge at rest has the maximum potential energy and zero kinetic energy. But when the positive charge is released to move in a uniform electric field, it has the same direction of motion as that of the electric field due to the electrostatic force acting on it.
As this positive charge moves in the direction of the uniform electric field, its velocity increases because according to the law of conservation of energy the initial potential energy of the positive charge gets converted into kinetic energy which provides energy for the motion of the charge.

Thus, the potential energy of the positive charge released in the uniform electric field decreases because this charge moves along the direction of the electric field.

Hence, the correct option is C.

Note: One can also use the formula for electric field which shows that as the distance travelled by the charge increases, the potential hence the potential energy of the charge decreases. One can also prove the same by proving that the work done by the electric field which should be equal to the negative of the change in potential energy of the charge is positive.