Two positive and equal charges are fixed at a certaindistance. A third...
Explanation:
When a small charge is placed in between two positive and equal charges, it experiences a net force due to the repulsion of the two charges. The direction of the net force depends on the position of the small charge relative to the two fixed charges.
a) If the small charge is positive, it will experience a net repulsive force from the two fixed charges. This force will push it away from the equilibrium position. Therefore, the equilibrium is not stable.
b) If the small charge is negative, it will experience a net attractive force from the two fixed charges. This force will pull it towards the equilibrium position. However, this equilibrium is also not stable as the small charge will continue to move towards one of the fixed charges and eventually stick to it.
c) The equilibrium is not always stable as explained in the above points.
d) The equilibrium is not stable because if the small charge is displaced slightly from the equilibrium position, it will experience a net force in one direction, causing it to move further away from the equilibrium position. This process will continue until the small charge either collides with one of the fixed charges or escapes to infinity.
Therefore, the correct answer is option 'D' - The equilibrium is not stable.
Two positive and equal charges are fixed at a certaindistance. A third...
Let the charges qA and qB which are positive and equal at a certain distance r
qC is the small charge placed at a distance r/2 b/w qA and qB
Since |qA|=|qB|
but qA and qB are in opposite direction net charge =0
hence, there is equilibrium established
but when qC with a small charge is bought at a distance r/2 b/W qA and qB
the equilibrium gets disturbed
Hence, option D is correct