Question

Q1) Two equal positive charge q are on the y axiz. One at y=+a and the other at y=-a

(a). Show that the electric field on the x-axis is along the x axis with Ex=2kqx(x^2+y^2) ^-3.2. Show that near the origin, when x is much smaller than a, Ex is approximately 2kqx/a^3.

(b). Show that for values of x much larger than a Ex=2kq/x^2. Why is this result something you should have been able to at least almost guess?

Q2. Draw a picture of the electric field lines for situation described in Q1.

Q3. At the origin (in the same situation as the previous questions) the electric field is zero. That means that the electric force is zero. That means if I put a charge there it will stay there.

(a). Discuss the stability of the equilibrium for a positive charge by considering small displacements from equilibrium along the x axis and then along the y axis.

(b). Repeat part (a) for a negative charge.

(c). Find,the magnitude and charge needed to exactly result in the net force on each of the three charges to be zero.

(d). What will happen if any of these charges (in c) is displaced slightly from equilibrium?

Q4. A bead of masses m and a charge -q is made to slide frictionlessly along a thread that is situated along the x axis.

(a). Show that for small displacements the bead manifests simple harmonic motion.

(b). Find the period of that motion.

Answer 1

1.

due to one charge is given by Ex = Ex/(a2 + x2)1/2. The total field at (x, 0) is therefore E = 2kqx/(a2 + x2)3/2 i

point is given by E = kq/(a2 + x2). By symmetry, the y components of the E fields cancel. The x component of E

(b) For x << a, (a2 + x2)3/2 @ a3, and Ex @ 2kqx/a3

(c) Similarly, for x >> a, Ex @ 2kq/x2. This is to be expected; for x >> a, the system looks like a single charge of q

3.

(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