a. Explain why, for any charge configuration, the equipotential s… - Vidyadip

Question

$a.$ Explain why, for any charge configuration, the equipotential surface through a point is normal to the electric field at that point.
Draw a sketch of equipotential surfaces due to a single charge $(-q)$, depicting the electric field lines due to the charge.
$b.$ Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side a as shown alongside.

✓

Answer

$a.$ The work done in moving a charge from one point to another on an equipotential surface is zero. If the field is not normal to an equipotential surface, it would have a non zero component along the surface. This would imply that work would have to be done to move a charge on the surface which is contradictory to the definition of equipotential surface. Mathematically
Work done to move a charge dq, on a surface, can be expressed as $d W=d q(\vec{E} \cdot \overrightarrow{d r})$
But $dW = 0$ on an equipotential surface
$\therefore \vec{E} \perp \overrightarrow{d r}$
Equipotential surfaces for a charge $-q$ is shown alongside.

$b.$ Work required to dissociate the system of three charges
$=- \text { P.E. of the system }$
$=-\frac{1}{4 \pi \varepsilon_0}\left[\frac{q \times(-4 q)}{a}+\frac{q \times 2 q}{a}+\frac{(-4 q) \times 2 q}{a}\right]$
$=-\frac{1}{4 \pi \varepsilon_0}\left[-4 q^2+2 q^2-8 q^2\right]=+\frac{10 q^2}{4 \pi \varepsilon_0 a}$

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