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The work done to move a charge along an equipotential from $\mathrm{A}$ to $\mathrm{B}$.
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is zero
is zero
Work done in displacing a charge particle is given by $\mathrm{W}_{12}=\mathrm{q}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ and the line integral of electrical field from point 1 to 2 gives potential difference $\left(\mathrm{V}_2-\mathrm{V}_1\right)$
$$
=-\int_1^2 \mathrm{E} d \mathrm{~d}
$$
The charge q is moving in the electric field then work done $w=-q \int_1^2(E) d l$
For the potential on equipotential surface, $\left(V_2-V_1\right)=0$. Hence, work done on moving a charge is zero $(\mathrm{w}=0)$.
$$
=-\int_1^2 \mathrm{E} d \mathrm{~d}
$$
The charge q is moving in the electric field then work done $w=-q \int_1^2(E) d l$
For the potential on equipotential surface, $\left(V_2-V_1\right)=0$. Hence, work done on moving a charge is zero $(\mathrm{w}=0)$.
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