Chapter 2 - Section 5.2 - Conductors - Problem - Page 102: 38

$a) $ $\sigma_b = \frac{+q}{4\pi b^2}$ $\sigma_a= \frac{-q}{4\pi a^2}$ $\sigma_R = \frac{+q}{4\pi R^2}$ $b) \space V_{(C)} = \frac{+q}{4\pi \epsilon_0} \big[\frac{1}{b} + \frac{1}{R} - \frac{1}{a}\big]$ $c) $ $\sigma_b = 0 $ $\sigma_a= \frac{-q}{4\pi a^2}$ $\sigma_R = \frac{+q}{4\pi R^2}$ $V_{(C)} = \frac{+q}{4\pi \epsilon_0} \big[\frac{1}{R} - \frac{1}{a}\big]$

Create a spherical guassian surface with radius $r$ such that $a\le r\le b$ Since electric field inside a conductor is zero. Hence the inner part of the shell with inner radius a will get induced with a charge $-q$ so to make net charge $0$ and hence there will be no electric field inside the shell. Now a positive charge $+q$ will get induced on the outer surface of shell with outer radius $b$. $a) $ $\sigma_b = \frac{+q}{4\pi b^2}$ $\sigma_a= \frac{-q}{4\pi a^2}$ $\sigma_R = \frac{+q}{4\pi R^2}$ $b) $ Let's denote the center of sphere with C so $V_{(C)} - V_{\infty} = -\int_{\infty}^0 \vec{E}\cdot \vec{dr}$ $V_{(C)} - 0 = -\big[\int_{\infty}^b \frac{+q}{4\pi \epsilon_0 r^2} dr + \int_b^a 0dr+\int_a^R \frac{+q}{4\pi \epsilon_0 r^2} dr + \int_R^0 0 dr\big]$ $V_{(C)} = -\big[\frac{+q}{4\pi \epsilon_0} \big(\frac{-1}{r}\big)_{\infty}^b + \frac{+q}{4\pi \epsilon_0} \big(\frac{-1}{r}\big)_a^R\big]$ $V_{(C)} = \frac{+q}{4\pi \epsilon_0} \big[\frac{1}{b} + \frac{1}{R} - \frac{1}{a}\big]$ c) If the outer surface of the shell is connected to group, So the potential of the outer sphere will turn to 0. Also since $V_{(R)} + V_{(a)} = 0$ so the potential due to V_{(b)} = 0 hence a charge $-q$ will be given to outer surface. Hence the final charge distribution would be $\sigma_b = 0 $ $\sigma_a= \frac{-q}{4\pi a^2}$ $\sigma_R = \frac{+q}{4\pi R^2}$ Regarding Potential $V_{(C)} - V_{\infty} = -\int_{\infty}^0 \vec{E}\cdot \vec{dr}$ $V_{(C)} - 0 = -\big[\int_{\infty}^b 0 dr + \int_b^a 0dr+\int_a^R \frac{+q}{4\pi \epsilon_0 r^2} dr + \int_R^0 0 dr\big]$ $V_{(C)} = -\big[ \frac{+q}{4\pi \epsilon_0} \big(\frac{-1}{r}\big)_a^R\big]$ $V_{(C)} = \frac{+q}{4\pi \epsilon_0} \big[\frac{1}{R} - \frac{1}{a}\big]$