Q#23.27
A thin spherical shell with radius $R_1$ = 3.00 cm is concentric with a larger thin spherical shell with radius $R_2$ = 5.00 cm. Both shells are made of insulating material. The smaller shell has charge $q_1$ = +6.00 nC distributed uniformly over its surface, and the larger shell has charge $q_2=-$ 9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells.
(a) What is the electric potential due to the two shells at the following distance from their common center: (i) r = 0; (ii) r = 4.00 cm; (iii) r = 6.00 cm?
(b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?
Answer:
The potential at any point is the scalar sum of the potential due to each shell.
$V=\frac{kq}{R}$ for r $\leq$ R and $V=\frac{kq}{r}$ for r > R.
(a) (i) r = 0. This point is inside both shells so
$V=k\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}\right)$
$V=(8.99 \times 10^9 \ N.m^2/C^2)\left(\frac{6.00 \times 10^{-9} \ C}{0.0300 \ m}+\frac{-9.00 \times 10^{-9} \ C}{0.0500 \ m}\right)$
$V=+1.798 \times 10^3 \ V + (-1.618 \times 10^3 \ V)=180$ V.
(ii) r = 4.00 cm. This point is outside shell 1 and inside shell 2.
$V=k\left(\frac{q_1}{r}+\frac{q_2}{R_2}\right)$
$V=(8.99 \times 10^9 \ N.m^2/C^2)\left(\frac{6.00 \times 10^{-9} \ C}{0.0400 \ m}+\frac{-9.00 \times 10^{-9} \ C}{0.0500 \ m}\right)$
$V=+1.348 \times 10^3 \ V + (-1.618 \times 10^3 \ V)=-270$ V.
(iii) r = 6.00 cm. This point is outside both shells
$V=k\left(\frac{q_1}{r}+\frac{q_2}{r}\right)=\frac{k}{r}(q_1+q_2)$
$V=\frac{8.99 \times 10^9 \ N.m^2/C^2}{0.0600 \ m}(6.00 \times 10^{-9} \ C + (-9 \times 10^{-9} \ C))$
$V=-145 \ V$
(b) At the surface of the inner shell, r =$R_1$ = 3.00 cm. This point is inside the larger shell, so
$V=k\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}\right)$
$V=(8.99 \times 10^9 \ N.m^2/C^2)\left(\frac{6.00 \times 10^{-9} \ C}{0.0300 \ m}+\frac{-9.00 \times 10^{-9} \ C}{0.0500 \ m}\right)$
$V=+1.798 \times 10^3 \ V + (-1.618 \times 10^3 \ V)=180$ V.
At the surface of the outer shell, r = $R_2$ = 5.00 cm. This point is outside the smaller shell, so
$V=k\left(\frac{q_1}{r}+\frac{q_2}{R_2}\right)$
$V=(8.99 \times 10^9 \ N.m^2/C^2)\left(\frac{6.00 \times 10^{-9} \ C}{0.0500 \ m}+\frac{-9.00 \times 10^{-9} \ C}{0.0500 \ m}\right)$
$V=+1.079 \times 10^3 \ V + (-1.618 \times 10^3 \ V)=-539$ V.
The potential difference is $V_1-V_2$ = 719 V. The inner shell is at higher potential. The potential difference is due entirely to the charge on the inner shell.
Inside a uniform spherical shell, the electric field is zero so the potential is constant (but not necessarily zero).
Q#23.28
A total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius of 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm, (b) 24.0 cm, (c) 12.0 cm.
Answer:
(a) This is outside the sphere, so
$V=\frac{kq}{r}$
$V=\frac{(8.99 \times 10^9 \ N.m^2/C^2)(3.50 \times 10^{-9} \ C)}{0.480 \ m}=65.6 \ V$
(b) This is at the surface of the sphere, so
$V=\frac{(8.99 \times 10^9 \ N.m^2/C^2)(3.50 \times 10^{-9} \ C)}{0.240 \ m}=131 \ V$
(c) This is inside the sphere. The potential has the same value as at the surface, 131 V.
All points of a conductor are at the same potential.
Q#23.29
A uniformly charged, thin ring has radius 15.0 cm and total charge +24.0 nC. An electron is placed on the ring’s axis a distance 30.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest.
(a) Describe the subsequent motion of the electron.
(b) Find the speed of the electron when it reaches the center of the ring.
Answer:
(a) The force on the electron exerted by this field is given by Eq. $\vec{E}=\frac{\vec{F}_0}{q_0}$.
When the electron is on either side of the center of the ring, the ring exerts an attractive force directed toward the center of the ring. This restoring force produces oscillatory motion of the electron along the axis of the ring, with amplitude 30.0 cm.
The force on the electron is not of the form F = –kx so the oscillatory motion is not simple harmonic motion.
(b) Apply conservation of energy to the motion of the electron.
$K_a+U_a=K_b+U_b$
with a at the initial position of the electron and b at the center of the ring.
We use, $V=\frac{1}{4\pi \epsilon_0}\frac{Q}{\sqrt{R^2+x^2}}$, where R is the radius of the ring.
$x_a$ = 30.0 cm, $x_b$ = 0, $K_a$ = 0 (released from rest) and $K_b=\frac{1}{2}mv^2$
$V=\frac{1}{4\pi \epsilon_0}\frac{Q}{\sqrt{R^2+x_a^2}}$
$V_a=(8.988 \times 10^9 \ N.m^2/C^2)\frac{24.0 \times 10^{-9} \ C}{\sqrt{(0.150 \ m)^2+(0.300 \ m)^2}}$
$V_a$ = 643 V
$V_b=\frac{1}{4\pi \epsilon_0}\frac{Q}{\sqrt{R^2+x_b^2}}$
$V_b=(8.988 \times 10^9 \ N.m^2/C^2)\frac{24.0 \times 10^{-9} \ C}{\sqrt{(0.150 \ m)^2+(0)^2}}$
$V_b$ = 1438 V
Thus $\frac{1}{2}mv^2=U_a-U_b$
And U = qV = $-eV$
$\frac{1}{2}mv^2=(-eV_a)-(-eV_b)$
$\frac{1}{2}mv^2=e(V_b-V_a)$ so
$v=\sqrt{\frac{2e(V_b-V_a)}{m}}$
$v=\sqrt{\frac{2(1.602 \times 10^{-19} \ C)(1438 \ V -643 \ V)}{9.109 \times 10^{-31} \ kg}}$
$v=1.67 \times 10^7$ m/s
The positively charged ring attracts the negatively charged electron and accelerates it. The electron has its maximum speed at this point. When the electron moves past the center of the ring the force on it is opposite to its motion and it slows down.
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