Question

Why must electrostatic fields at the surface of a charged conductor be normal to the surface at every point? Give reason.

Answer 1

Hint: Electric field is defined as the gradient of potential. This shows that the electric field depends on the potential of the conductor. The potential on the surface of a charged conductor is constant.
Formula used:
$E \propto \dfrac{F}{q}$
$E = - gradV$

Complete step-by-step solution:
The electric field is defined as the electric force per unit charge:
 $E \propto \dfrac{F}{q}$
Here, F is the force exerted or experienced by the object and q is the charge on the object. So, if the electric field at a particular point is known and force is known we can get the charge or if the charge is known we can get the force experienced by the point charge. The direction of the field is taken to be the direction of the force it would exert on the positive test charge.

Also, the Electric field is defined as the gradient of potential (V):
$E = - gradV$
Since we know that the surface of the charged conductor has a constant potential. Therefore, there is no field along the surface of the charged conductor.
This implies that the electrostatic field at the surface of the charged conductor should be normal to the surface at every point.

Additional information:
The electric field is radially outward from a positive charge and it is radially inward to a negative point charge. S.I unit of the electric field is Newton per coulomb.
Potential (V) is defined as the amount of work needed to move one unit of electric charge from a reference point to a specific point in an electric field.
S.I unit of potential is Volts or joules per coulomb.

Note: Electric field is a vector quantity. It has both magnitude and direction.
The electric field is not negative; the negative sign shows the opposite direction of the field.
The electric field inside a conductor is zero because charges are not present inside the conductor, they are present on the surface of the conductor.