Two equal positive point charges +q are on the y axis; one is at y = +a and the

Question

Two equal positive point charges +q are on the y axis; one is at y = +a and the other is at y = -a. The electric field at the origin is zero, A test charge q_0 placed atht e origin will therefore be in equilibrium. a) Discuss the stability of the equilibrium for a positive test charge by considering small displacements from equilibrium along the x axis and small displacements along the y axis. b) Repeat part (a) for a negative test charge. c) Find the magnitude and sign of a charge q_0 that when placed at the origin results in a net force of zero on each of the three charges. Two equal positive point charges +q are on the y axis; one is at y = +a and the other is at y = -a. The electric field at the origin is zero, A test charge q_0 placed atht e origin will therefore be in equilibrium. a) Discuss the stability of the equilibrium for a positive test charge by considering small displacements from equilibrium along the x axis and small displacements along the y axis. b) Repeat part (a) for a negative test charge. c) Find the magnitude and sign of a charge q_0 that when placed at the origin results in a net force of zero on each of the three charges.

Explanation / Answer

(a) Since Ex is in the x direction, a positive test charge that is displaced from (0, 0) in the x direction will experience a force in the x diretion and accelerate in the x direction. Consequently, the equilibrium at (0, 0) is unstable for a small displacement along the x axis. If the positive test charge is displaced in the y direction, the charge at +a will exert a greater force than the charge at –a, and the net force is then in the –y direction; i.e., it is a restoring force. Consequently, the equilibrium at (0, 0) is stable for small displacements along the y direction. (b) Following the same arguments as in part (a), one finds that, for a negative test charge, the equilibrium is stable at (0, 0) for displacements along the x direction and unstable for displacements along the y direction. (c) Since the two +q charges repel, the charge Q at (0, 0) must be a negative charge. Since the force between charges varies as 1/r2, and the negative charge is midway between the two positive charges, Q = –q/4.

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