Question

A test charge q_{0} is slowly moved from point A to B perpendicular to the uniform electric field as shown in figure. The work done by the agent to move the charge fron to B against the electric field will be point A B E q q_{0} A q_{0}*E * d; (- q_{0} * E * d)/2 Zero​

Answer 1

Given:

A test charge q_{0} is slowly moved from point A to B perpendicular to the uniform electric field as shown in figure.

To find:

Work done by agent ?

Calculation:

The work done in moving the charge be W :

$\rm\therefore W = \vec{F} \: . \: \vec{d}$

• F - Force and d - Displacement

$\rm\implies W = q_{0}( \vec{E}) \: . \: \vec{d}$

• Now, as per the question, the angle between the field intensity vector and the displacement vector is 90°.

$\rm\implies W = q_{0} \times E \times d \times \cos( {90}^{ \circ} )$

$\rm\implies W = q_{0} \times E \times d \times 0$

$\rm\implies W = 0$

So , work done by agent is zero (0).