The displacement of a charge $Q$ in the electric field $\vec E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k$ is $\vec r = a\hat i + b\hat j$. The work done is

$Q\sqrt {{{\left( {a{e_1}} \right)}^2} + {{\left( {b{e_2}} \right)}^2}} $

$Q\left( {{e_1} + {e_2}} \right)\sqrt {{a^2} + {b^2}} $

$Q\left( {\sqrt {e_1^2 + e_2^2} } \right)\left( {a + b} \right)$

JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Application +4 | -1

The displacement of a charge

Q

in the electric field

\vec E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k

\vec r = a\hat i + b\hat j

. The work done is

View answer and explanation

Correct option: A

Q\left( {a{e_1} + b{e_2}} \right)

Option A (

Q\left( {a{e_1} + b{e_2}} \right)

) is correct because the work done is the dot product of the force (Q*E) and the displacement vector.

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