Question

(a) Explain why, for a 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 in the figure.

Answer 1

a) As we know work done in moving a test charge along an equipotential surface is zero and this is because an equipotential surface is a surface with a constant value of potential at all points on the surface.

For non-zero displacement this is possible thus, the force acting on the point charge is perpendicular to the equipotential surface.

We also know that the lines of force or the electric field lines indicate the direction of electric force on a charge, therefore, for any charge configuration, equipotential surface through a point is normal to the electric field.

b) Work done in dissociate this charge configuration will be negative of the potential energy of the system.

If A, B, C are the corners then,

U= - (Charge on corner A/AB+ charge on corner B/BC+ charge on corner C/AC)

Work done can be calculate by using the above equation.

$⟹U=\frac{1}{4\pi ϵ}[\frac{q(-4q)}{a}+\frac{(-4q)\left(2q\right)}{a}+\frac{\left(2q\right)\left(q\right)}{a}]=\frac{1}{4\pi ϵ}\left[\frac{(-10q^2)}{a}\right]$