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Question
A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge ______.
Options
remains a constant because the electric field is uniform
increases because the charge moves along the electric field
decreases because the charge moves along the electric field
decreases because the charge moves opposite to the electric field
Solution
A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge decreases because the charge moves along the electric field.
Explanation:
Electric potential decreases in the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at the high electrostatic potential to another equipotential surface maintained at low electrostatic potential.
The positively charged particle experiences electrostatic force along the direction of electric field and hence moves in the direction of electric field. Thus, positive work is done by the electric field on the charge. We know
`W_("electrical") = -ΔU = -qΔV = q(V_("initial") – V_("final"))`
Hence electrostatic potential energy of the positive charge decreases.
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