Solutions to Assignment 3

FGT 24.29

The electric field is given by, $\vec{E}=bx^2\hat{i}$. We seek the electric flux, $\phi_E$, through the surfaces of a cube which has one corner at the origin, which has its base in the x-y plane, and which extends from 0 to a on each of the x, y and z axes. We also would like to know what charge is enclosed in the box.

Solution

Since the electric flux is in the $\hat{i}$ direction, the only finite flux is due to the surfaces with normals which are parallel to $\hat{i}$. The other surfaces have normals which lie along the $\hat{j}$ or $\hat{k}$ directions and so the dot product $\vec{E}\cdot \vec{dA} = \vec{E}\cdot \hat{n}dA$is zero in those cases. In addition the surface which is at x=0 has zero flux, as there the electric field magnitude is zero. The only surface which contributes a finite flux is the surface oriented in the $\hat{i}$ direction and which lies at x=a. The flux through this surface is,

$$\phi_E (x=a) = E(a). Area = ba^4$$ (1)

From Gauss' law, we have $\phi_E = q/\epsilon_0$, so we find,

$$q = \epsilon \phi_e = \epsilon_0 ba^4$$ (2)

FGT 24.52

There is a charge density of $\sigma_1 = 30\mu C/m^2$ on the surface of a conducting sphere of radius r₁ = 0.3m. In addition, there is a uniform charge density of $\sigma_2=15\mu C/m^2$on the surface of a spherical shell of radius r₂ = 0.8m. a) Find the electric field at a = 0.5m; b) Find the electric field at b=0.7m; c) How would the answers to a) and b) change if the outer shell of charge were removed ; d) Find the electric field at r=1.

Solution

a) From the shell theorem, the outer shell of radius r₂gives no contribution to the electric field at a<r₂. Therefore we just consider the inner charge. The inner charge is $q_1 = 4\pi r_1^2 \sigma_1$. The electric field at a is just the electric field due to this point charge located at the center of the system. We thus find that the magnitude of the electric field at a is,

$$E_a = {k q_1 \over a^2} = {4\pi r_1^2 \sigma_1 \over 4\pi \epsilon_0 a^2} = {\sigma_1 r_1^2\over \epsilon_0 a^2}$$ (3)

Plugging the numbers gives, $E_a=1.221\times 10^6N/C$

b) Since we are still at a radius b<r₂ it is only the inner charge that contributes, we thus have the same formula, but with $a\rightarrow b$, ie.

$$E_b = {\sigma_1 r_1^2\over \epsilon_0 b^2}$$ (4)

Plugging the numbers gives, $E_b = 6.23\times 10^5N/C$.

c) The solutions to a) and b) are not changed if we remove the outer shell of charge, as it gives zero electric field contribution in its interior.

d) The electric field at r>b has contributions from both shells of charge. The shell theorem holds, so the electric field at r=1 is like that of a point charge of size $q_1+q_2 = 4\pi (r_1^2\sigma_1 + r_2^2 \sigma_2)$We thus have,

$$E_r = {\sigma_1 r_1^2 + \sigma_2 r_2^2\over \epsilon_0 r^2}$$ (5)

Plugging the numbers gives, $E_r = 1.39 \times 10^6N/C$.

24.55

You interpretted this question in two different ways. The answer in the back of the book is incorrect for either interpretation. There is a subtlety in this problem that FGT seems to have missed and which no-one got right. The subtlety is that if we have a non-uniform charge density, $\rho(R)$, the charge inside a sphere of charge is given by,

$$q(r) = \int_0^r \rho(R) dV(R)$$ (6)

of course this reduces to $\rho V$ if the charge density is constant, but in general we have to do the integral.

Solution - First interpretation, the "right'' one

Since the question asks for a charge distribution which leads to an electric field which is directed radially outward from an axis, we assume a cylindrical symmetry. From Gauss' law, the electric field and the enclosed charge are related by,

$$E 2\pi r L = {1\over \epsilon_0} L \int_0^r 2\pi R \rho(R) dR$$ (7)

In order that the electric field be a constant we need the integral on the RHS to be proportional to r. For this to occur, we need the charge density to be $\rho(R) = c/R$, where c is a constant. If we assume this form, we can do the integral and find that $c= \epsilon_0E$. We thus have,

$$\rho(r) = {\epsilon_0 E \over r}$$ (8)

Solution - Second interpretation

If we assume that the question refers to a spherical situation, where we seek a charge distribution which will produce a uniform radial field, then we have,

$$E 4\pi r^2 = {1\over \epsilon_0} \int_0^r 4\pi R^2 \rho(R) dR$$ (9)

Again we assume a charge density $\rho(r) = d/r$ where d is a different constant. In this case, we find, $d=2\epsilon_0 E$, so that

$$\rho(r) = {2\epsilon_0 E\over r}$$ (10)

FG24.58

We consider a uniform sphere of constant charge density $\rho$, and of radius R. a) Find the electric field for r<R. b) Now cut a small sphere out of size b<<R out of this sphere of charge. The center of this small cavity is located at $\vec{a}$. Find the electric field inside this small spherical cavity.

Solution

a) For r<R, the electric field is given by Gauss' theorem which gives,

$$4\pi r^2 E(r) = {q(r) \over \epsilon_0} = {1\over \epsilon_0} {4\pi r^3 \rho \over 3}$$ (11)

From this we find that,

$$\vec{E} = {\rho r \hat{r} \over 3\epsilon_0} = {\rho \vec{r} \over 3\epsilon_0}$$ (12)

b) In order to calculate the electric field inside the small spherical cavity, we use superposition. We consider the original sphere of positive charge and add to it a small uniform sphere of negative charge density, $-\rho$centered at $\vec{a}$ and of radius b. The electric field inside the cavity due to the contributions of these two spheres of charge is given by the vector sum,

$$\vec{E}_{tot} = {\rho \vec{r} \over 3\epsilon_0} - {\rho \vec{r}_a \over 3\epsilon_0}$$ (13)

where $\vec{r}_a$ is the radial vector whose origin is the center of the small sphere of negative charge. Now note that we can relate $\vec{r}_a$ and $\vec{r}$ by,

$$\vec{r} = \vec{a} + \vec{r}_a$$ (14)

We thus find the simple result that the electric field inside the cavity is uniform and is given by,

$$\vec{E}_{tot} = {\rho \vec{r} \over 3\epsilon_0} - {\rho \vec{r}_a \over 3\epsilon_0} = {\rho \vec{a} \over 3\epsilon_0}$$ (15)