(a) Consider an arbitrary electrostatic field configuration. A smalltest charge is placed at a null point (i.e., where E = 0) of theconfiguration. Show that the equilibrium of the test charge isnecessarily unstable. (b) Verify this result for the simple configuration of two charges ofthe same magnitude and sign placed a certain distance apart.
a)
Assume that a test charge is at stable equilibrium. When this test charge is displaced from its position then a restoring force is experienced towards the null point where electric field is zero. It indicates that there is a net inward flux through a closed surface around the null point. This statement contradicts Gauss’ Law which tells that flux through a closed surface not enclosing any charge should be zero. Therefore, the test charge must be unstable.
b)
A test charge in stable equilibrium continues to remain in equilibrium even when it is disturbed from its position. In the given case, the mid-point of the joining line is the null point. When the test charge is given some disturbance along its line then it experiences a restoring force towards mid-point, but when it is displaced in perpendicular direction then the net force pulls it away from the null point; therefore, it is unstable.
a)
Assume that a test charge is at stable equilibrium. When this test charge is displaced from its position then a restoring force is experienced towards the null point where electric field is zero. It indicates that there is a net inward flux through a closed surface around the null point. This statement contradicts Gauss’ Law which tells that flux through a closed surface not enclosing any charge should be zero. Therefore, the test charge must be unstable.
b)
A test charge in stable equilibrium continues to remain in equilibrium even when it is disturbed from its position. In the given case, the mid-point of the joining line is the null point. When the test charge is given some disturbance along its line then it experiences a restoring force towards mid-point, but when it is displaced in perpendicular direction then the net force pulls it away from the null point; therefore, it is unstable.
