The charge enclosed by the gaussian surface is rpr2l.
Apply Gauss's law
ò EdA = Q/e0
òbase EdA + òcurvature EdA = 0 + E2prl = rpr2l/e0
which gives E = (rr/2e0)r.
QUIZ 2
1. Consider a long cylindrical charge distribution of radius R with a uniform charge density r. Find the electric field at distance r from the axis where r < R.
| Answer. Choose a cylindrical gaussian surface
with radius r and length l centered at the axis of the
charge distribution as shown.
The charge enclosed by the gaussian surface is rpr2l. Apply Gauss's law ò EdA = Q/e0 òbase EdA + òcurvature EdA = 0 + E2prl = rpr2l/e0 which gives E = (rr/2e0)r. |
|
2. Two point charges, Q1 = +5.00nC and Q = -3.00nC, are separated by 35.0cm. (a) Whhat is the potential energy of the pair? (b) What is the electric potential at a point midway between the charges?
Answer.
(a) The electric potential of the pair can be calculated as one charge's energy due to the other's potential, the potential at Q1 due to Q2 is
V1 = kQ2/r = (9.0 x 109C2/Nm2)(-3.00 x 10-9C)/(35.0 x 10-2m) = -77.14V
The potential energy of Q1 is
U1 = V1Q1 = (-77.14V)(5.00 x 10-9C) = -3.86 x 10-7J
Therefore the potential energy of the pair is -3.86 x 10-7J = -386nJ.
(b) At the point midway between the charge, the potential is the sum of the potentials due to two charges separately.
V = V1 + V2 = (9.0 x 109C2/Nm2)(+5.00 x 10-9C)/(35.0/2 x10-2m) + (9.0 x10-9C2/Nm2)(-3.00 x10-9C)/(35.0/2 x10-2m)
= 103V