A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge.

Options

A. remains constant because the electric field is uniform.
B. increases because the charge moves along the electric field.
C. decreases because the charge moves along the electric field.
D. decreases because the charge moves opposite to the electric field.

Answer

C

Explanation

When a positively charged particle is released in a uniform electric field, it moves in the direction of the electric field. The relation between the magnitude of the electric field intensity (E ) and the electric potential (V) is given by, `E=(dV)/(dr)`, By the above relation we can conclude that the electric potential decreases in the direction of the electric field. The electric potential energy of charged particle is given by,
`U=qV`
As V decreases on moving along the direction of the electric field thus, the electric potential energy (U) will also decrease as the charge moves in the direction of the electric field.