Solution

$E = \dfrac{kQx}{(a^2 + x^2)^{\frac{3}{2}}} \hat i$

Solution

$E = \frac{\sigma}{2\epsilon_0}$
A typical ring has charge $dQ$, inner radius $r$, and outer radius $r + dr$. Its area is approximately equal to its width $dr$ times its circumference $2\pi r$, or $dA = 2\pi r dr$. The charge per unit area is $\sigma = dQ/dA$, so the charge of the ring is $dQ = \sigma dA = 2\pi \sigma r dr$.
The expression for the field due to a ring that we found in example 21.9,

$\begin{aligned} dE_x &= \frac{1}{4\pi \epsilon_0} \frac{2\pi \sigma r x dr}{(x^2 + r^2)^{3/2}} \end{aligned}$

To find the total field due to all the rings, we integrate $dE_x$, over $r$ from $r=0$ to $r=R$ (not from $-R$ to $R$):

$\begin{aligned} E_x &= \int_0^R \frac{1}{4\pi \epsilon_0} \frac{2\pi \sigma r x dr}{(x^2 + r^2)^{3/2}}\\ &= \frac{\sigma x}{4 \epsilon_0} \int_0^R \frac{2 r dr}{(x^2 + r^2)^{3/2}}\\ &= \frac{\sigma x}{2 \epsilon_0} \lb -\frac{1}{\sqrt{x^2 + R^2} + \frac{1}{x}} \rb\\ &= \frac{\sigma}{2 \epsilon_0} \lb 1 -\frac{1}{\sqrt{R^2/x^2 + 1}} \rb \quad \bold{(21.11)} \end{aligned}$

If the disk is very large (or if we are close to it), so that $R \gg x$, the term $\frac{1}{\sqrt{R^2/x^2 + 1}}$ is very much less than 1. Then Eq. (21.11) becomes

$\begin{aligned} E = \frac{\sigma}{2\epsilon_0} \quad \bold{(21.12)} \end{aligned}$

Our final result does not contain the distance $x$ from the plane. Hence the electric field produced by an infinite plane sheet of charge is independent of the distance from the sheet. The field direction is everywhere perpendicular to the sheet, away from it. There is no such thing as an infinite sheet of charge, but if the dimensions of the sheet are much larger than the distance $x$ of the field point $P$ from the sheet, the field is very nearly given by Eq. (21.12)

Solution

$\begin{aligned} \vec E = \vec E_1 + \vec E_2 = \begin{cases} 0 &\text{Above the upper sheet}\\ \frac{\sigma}{\epsilon_0}\jhat &\text{Between the sheets}\\ 0 &\text{Below the upper sheet} \end{cases} \end{aligned}$

Solution

a. Let the charge density be $\lm = \dfrac{Q}{a}$. At point x, the y-component of electric field is zero, and the x-component of electric field is the electri field produced by $dQ$ of the charge distribution Q.

$\begin{aligned} dQ &= \lm dx\\ dE_x &= \frac{k dQ}{(a+r-x)^2} = \frac{k \lm dx}{(a+r-x)^2}\\ E_x &= k \lm \int_0^a \frac{dx}{(a+r-x)^2}\\ &= k \lm \bigg(\frac{1}{a+r-x}\bigg) \bigg\vert_0^a\\ &= \frac{k \lm \cdot a}{r(a+r)}\\ &= \frac{k Q}{r(a+r)}\\ E &= E_x \hat i = \frac{k Q}{r(a+r)} \hat i \end{aligned}$

b. The electric force $\vec F = Eq = \dfrac{k Qq}{r(a+r)} \hat i$. The direction is +x-axis.
c. When $r \gg a$,

$\begin{aligned} r(a+r) &\approx r^2\\ F &= \frac{k Qq}{r(a+r)} \\ &=\frac{k Qq}{r^2} = \frac{Qq}{4\pi \epsilon_0 r^2} \end{aligned}$

Solution

Let the charge density be $\lm = Q/a$, we have $dQ = \lm \cdot dy$
The x-component of the electric field is

$\begin{aligned} dE_x &= \frac{kdQ}{r^2} \cos \th\\ &= k \lm dy \cdot \frac{1}{x^2 + y^2} \cdot \frac{x}{\sqrt{x^2 + y^2}}\\ &= k \lm x\cdot \frac{1}{(x^2 + y^2)^{\frac{3}{2}}}\cdot dy\\ E_x &= k\lm x \int_0^a \frac{1}{(x^2 + y^2)^{\frac{3}{2}}}dy\\ &= \frac{kQx}{a} \big(\frac{y}{x^2\sqrt{x^2 + y^2}}\big)\big\vert_0^a\\ &= \frac{kQ}{x\sqrt{x^2 + a^2}} \end{aligned}$

The y-component of the electric field is

$\begin{aligned} dE_y &= \frac{kdQ}{r^2} \sin \th\\ &= k \lm dy \cdot \frac{1}{x^2 + y^2} \cdot \frac{y}{\sqrt{x^2 + y^2}}\\ &= k \lm \cdot \frac{y}{(x^2 + y^2)^{\frac{3}{2}}}\cdot dy\\ E_y &= k\lm \int_0^a \frac{y}{(x^2 + y^2)^{\frac{3}{2}}}dy\\ &= \frac{kQ}{a} \big(-\frac{1}{\sqrt{x^2 + y^2}}\big)\big\vert_0^a\\ &= \frac{kQ}{a} \big( \frac{1}{x} - \frac{1}{\sqrt{x^2 + a^2}}\big) \end{aligned}$

c. Hint: $(1 + z)^n = 1 + nz + \cdots$

$\begin{aligned} \frac{1}{\sqrt{x^2 + a^2}} &= \frac{1}{x} (1-\frac{a^2}{2x^2})\\ F_x = - E_x q &= \frac{kQ}{x\sqrt{x^2 + a^2}}\\ &=-\frac{kQ}{x^2}(1-\underbrace{\frac{a^2}{2x^2}}_{\blue{approaches 0}})\\ &=-\frac{kQ}{x^2} = \frac{-Qq}{4\pi\epsilon_0 x^2}\\ F_y = E_y q &= \frac{kQq}{a} \big( \frac{1}{x} - \frac{1}{\sqrt{x^2 + a^2}}\big)\\ &= \frac{kQ}{a} \cdot \frac{1}{x} \big(1 - 1 + \frac{a^2}{2x^2}\big)\\ &= \frac{kQa}{2x^3} =\frac{+Qqa}{8\pi\epsilon_0 x^3} \end{aligned}$

Solution

Let the magnitude of the charge density be $\d\lm = \frac{Q}{a\cdot \frac{\pi}{2}} = \frac{2Q}{\pi a}$.

$dQ = \lm \cdot a d\th = \frac{2Q}{\pi}d\th$

The x-component of the electri field is

$\begin{aligned} dE_x &= \frac{k dQ}{a^2} \cos \th\\ E_x &= \frac{k}{a^2} \cdot \frac{2Q}{\pi} \int_0^{\pi/2} \cos \th d\th\\ &= \frac{2kQ}{\pi a^2} (\sin \th) \big\vert_0^{\pi/2}\\ &= \frac{2kQ}{\pi a^2} \end{aligned}$

The y-component of the electri field is

$\begin{aligned} dE_y &= \frac{k dQ}{a^2} \sin \th\\ E_y &= \frac{k}{a^2} \cdot \frac{2Q}{\pi} \int_0^{\pi/2} \sin \th d\th\\ &= \frac{2kQ}{\pi a^2} (-\cos \th) \big\vert_0^{\pi/2}\\ &= \frac{2kQ}{\pi a^2} \end{aligned}$

Solution

Let the electric field of positive charge be $E_1$, and that of negative charge be $E_2$. From (21.85), we know that

$\begin{aligned} E_1 &= \frac{2kQ}{\pi a^2} \ihat -\frac{2kQ}{\pi a^2} \jhat\\ E_2 &= \frac{2kQ}{\pi a^2} \ihat +\frac{2kQ}{\pi a^2} \jhat\\ E_{total} &= E_1 + E_2\\ &= \frac{4kQ}{\pi a^2} \ihat \end{aligned}$

Solution

The electric field produced by a uniformly charged sheet at any distance is given by

$\begin{aligned} E = \frac{\sigma}{2\epsilon_0} \end{aligned}$

Sheet A gives an electric field of

$\begin{aligned} E_A &= \frac{\sigma}{2\epsilon_0} = \frac{-8.8 \times 10^{-6}}{2\cdot 8.85 \times 10^{-12}}\\ &= -4.97 \times 10^5 N/C \end{aligned}$

where the negative sign means the field points towards sheet A, in any point of the space.
The electric field produced by sheet B is given by

$\begin{aligned} E_B &= \frac{\sigma}{2\epsilon_0} = \frac{-11.6 \times 10^{-6}}{2\cdot 8.85 \times 10^{-12}}\\ &= -6.55 \times 10^5 N/C \end{aligned}$

and again, the negative sign means that the field at any point of the space points towards sheet B.
a. 4.00 cm to the right of sheet A (between the two sheets)

$\begin{aligned} E &= E_A - E_B = -4.97 \times 10^5 N/C - (-6.55 \times 10^5 N/C)\\ &= 1.58 \times 10^5 N/C \end{aligned}$

And the direction is towards sheet B.
b. 4.00 cm to the left of sheet A

$\begin{aligned} E &= E_A + E_B = -4.97 \times 10^5 N/C + (-6.55 \times 10^5 N/C)\\ &= -11.52\times 10^5 N/C \end{aligned}$

towards right.
c. 4.00 cm to the right of sheet B.

$\begin{aligned} E &= E_A + E_B = -4.97 \times 10^5 N/C + (-6.55 \times 10^5 N/C)\\ &= -11.52\times 10^5 N/C \end{aligned}$

toward left.