In an electric field due to charge Q,a charge | vedclass

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(D) The electric field $\vec{E}$ due to a point charge $Q$ is always directed radially outward.Since the path from $A$ to $B$ is an arc of a circle ce

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(D) The electric field $\vec{E}$ due to a point charge $Q$ is always directed radially outward.Since the path from $A$ to $B$ is an arc of a circle centered at $Q$,the displacement vector $d\vec{s}$ at any point on the path is tangent to the circle.The radial direction (direction of $\vec{E}$) is always perpendicular to the tangent of the circle at any point.Therefore,the force $\vec{F} = q\vec{E}$ is always perpendicular to the displacement $d\vec{s}$.The work done $W$ is given by the integral $W = \int_A^B \vec{F} \cdot d\vec{s}$.Since $\vec{F} \perp d\vec{s}$,the dot product $\vec{F} \cdot d\vec{s} = F \cdot ds \cos 90^{\circ} = 0$.Thus,the total work done $W = 0$.

In an electric field due to charge $Q$,a charge $q$ moves from point $A$ to $B$ along an arc of a circle centered at $Q$,as shown in the figure. The work done is ($\varepsilon_0=$ permittivity of free space).

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