HW 10

9-9

For an infinite sheet of charges with a density $\sigma$, the electric field can be calculated using Gauss' law, by taking a Gaussian surface of area $A$ and thickness $d$ that encloses an area $A$ out of the sheet. Then the total flux is $\Phi = 2AE= \sigma A/\epsilon$, independent of $d$. So $E=\sigma/2\epsilon_0$ outwards. For two such sheets of charges, we just have to superimpose them. (a) To the left, and (c) to the right of the sheets, the electric field is zero. (b) in between, the electric field is from left to right, with a magnitude $E=\sigma/\epsilon_0$.

10-1

The electric potential difference is $\Delta V= 10^9~\rm V$, and the charge transferred is $Q=30~\rm C$. (a) The released energy is $E=\Delta VQ = 3.0\times 10^{10}~\rm J$. (b) From the kinetic energy equation ${1\over 2} mv^2 = E$, we get $v = \sqrt{2E\over m}=7.7\times 10^3 ~\rm m/s$. (c) From $L_f = 3.3\times 10^5 ~\rm J/kg$, $\Delta m = E/L_f = 9.1\times 10^4 ~\rm kg$.

10-2

From $e\Delta V= {1\over 2}m_ev^2$, we get $\Delta V = {m_ev^2 \over 2e} =0.50~\rm V$.

10-3

Let $d= 0.12~\rm m$ and $F = 3.9\times 10^{-15} ~\rm N$. (a) The electric field is $E=F/e = 2.4\times 10^4 ~\rm N/C$. (b) The potential difference is $\Delta V = Ed = 2.9\times 10^3 ~\rm V$.

10-4

The electric field vector is ${\bf E} = -\hat{\bf y} E_0$ with $E_0 = 325 ~\rm V/m$. The potential difference $ACB$ is $\Delta V = -\int_A^C {\bf E}\cdot d{\bf s}-\int_C^B {\bf E}\cdot d{\bf s} =2.6\times 10^2 ~\rm V$.

10-5

Since the electric field acting on the origin is the sum of the two with equal magnitude and in the opposite direction, it is zero. Consequently, the force is zero, too. (a) 0. (b) 0. (c) The electric potential is again the sum of the two: $V= 2Q / 4\pi \epsilon_0 r = 4.50\times 10^{4} ~\rm V$.

10-6

(a) $E_1 = -e^2/4\pi\epsilon_0 a_0 = -4.35\times 10^{-18} ~\rm J=-27.2~eV$. (b) $E_1 = -e^2/4\pi\epsilon_04 a_0 = -1.09\times 10^{-18} ~\rm J=-6.8~eV$. (c) $0$.

10-7

The radius of the grains is $R= 10^{-6}~\rm m$, and let the total charge $Q$ be the unknown number. Then the electric potential at the surface of the grains is $V= {Q\over 4\pi \epsilon_0 R}= -400 ~\rm V$. Then $Q= -4\times 10^{-14} {~\rm C} =3\times 10^5 e$.

10-8

From the equation of kinetic energy and the potential energy, ${1\over 2} m_e v^2 = {1\over 4\pi \epsilon_0} {eQ\over r}$, we get $v = \sqrt{eQ\over 2\pi \epsilon_0 m_e r} = 2.2\times 10^4 ~\rm m/s$.

10-9

(a) You need to plot the function $V(x/a) = {2Q \over 4\pi\epsilon_0 \sqrt{a^2+x^2}}$. (b) You need to plot the function $V(y/a) = {Q\over 4\pi\epsilon_0 a}({1\over \vert y/a-1\vert}-{1\over \vert y/a+1\vert})$.