# AP Physics Question. Gauss law

The figure below is a section of a conducting rod of radius $R1 = 1.30 mm$ and length $L = 11.00 m$ inside a thick-walled coaxial conducting cylindrical shell of radius $R2 = 10.0R1$ and the (same) length L. The net charge on the rod is $Q1 = +2.00*10^{-12}C$ that on the shell is $Q2 = 4.00Q1$.

Answer

a) $int_{interior-surface} (E*dS) = (Q1+Q2)/epsilon_0$

taking the interior surface a cylinder of radius r we have

$S =2*pi*r*L =3*pi*R_2*L$

$E = (Q_1+Q_2)/(3*pi*R2*L*epsilon_0) = 4*K_e*(Q_1+Q_2)/(3*R_2*L) = 0.839 (V/m)$

$K_e =9*10^9$

b)

lines of field emerge from + charges and enters – charges. It means lines are from the interior to the exteriorÂ (surface interior charge is positive)

c)

$E*S = (Q_1)/epsilon_0$

$S = 2*pi*r*L =11*pi*R_1*L$

$E = (Q_1)/(11*pi*R1*L*epsilon_0) =4*K_e*(Q_1)/(11*R_1*L) =0.4577 (V/m)$

d)

The same convention of sign: lines of field are from interior to exterior.

e)

Interior surface charge on the shell is zero. (If there were charges on the interior surface they would be moved by the nonzero field). (see also the Faraday cage)

f)

Exterior surface charge on the shell is $Q_2$