Question

You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using electric fields. The physical set up is shown in the picture below.

in the for a Fixed beads Movable bead +ng Three fixed beads are secured along the y-axis, with a bead of charge +ng at the origin and charges of -mq a distance a above and below the origin. The values of m and n can be set by an operator On an insulating wire laid along the x-axis, a movable bead with charge-pg is placed. By adjusting m and n, the movable bead can be moved to different equilbrium positions x along the wire. Your supervisor asks you to set up the system and test it. In particular, he asks for answers to the following questions. (a) Ifn 9 in a particular test, what is the value of m that will place the movable bead in equilibrium atx- a? you decide ter eress veur superver byfr ding the rapt of values of eve that will place the bead in equilibrium somewhere along the x axis. Note that m, n, and ρ are not necessarily integers. (Give your answer as an inequality. Use the folowing as necessary: m, n, p.) (d

Answer 1

(a) $x=a$

To keep the charge $-pq$ at equilibrium, net force on charge $-pq$ due to the 3 charges is zero.

Force on $-pq$ due to $+nq$  is  $\bold{F}_3=\frac{knpq^2}{a^2}(-\hat{i})$ is acting towards left.

Force on $-pq$ due to $-mq$ (upper) is $\bold{F}_1=\frac{kmpq^2}{(\sqrt{2}a)^2}(\cos45\degree\hat{i}-\sin45\degree\hat{j})$ and due to$-mq$ (lower)   is  $\bold{F}_2=\frac{kmpq^2}{(\sqrt{2}a)^2}(\cos45\degree\hat{i}+\sin45\degree\hat{j})$

Net force on  $-pq$ is zero, $\bold{F}_1+\bold{F}_2+\bold{F}_3=0$

$\frac{kmpq^2}{(\sqrt{2}a)^2}(\cos45\degree\hat{i}-\sin45\degree\hat{j})+\frac{kmpq^2}{(\sqrt{2}a)^2}(\cos45\degree\hat{i}+\sin45\degree\hat{j})-\frac{knpq^2}{(a)^2}\hat{i}=0$

$\frac{2kmpq^2}{(\sqrt{2}a)^2}(\cos45\degree\hat{i})-\frac{knpq^2}{(a)^2}\hat{i}=0$

$m\cos45\degree-n=0$

$m=n\sqrt{2}=9\sqrt{2}$

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(b) $x=2a$

To keep the charge $-pq$ at equilibrium, net force on charge $-pq$ due to the 3 charges is zero.

Force on $-pq$ due to $+nq$  is  $\bold{F}_3=\frac{knpq^2}{(2a)^2}(-\hat{i})$ is acting towards left.

Angle made by $-pq$ with $-mq$ is $\theta=\tan^{-1}\left ( \frac{a}{2a} \right )=26.6\degree$

Force on $-pq$ due to $-mq$ (upper) is $\bold{F}_1=\frac{kmpq^2}{(\sqrt{5}a)^2}(\cos26.6\degree\hat{i}-\sin26.6\degree\hat{j})$ and due to$-mq$ (lower)   is  $\bold{F}_2=\frac{kmpq^2}{(\sqrt{2}a)^2}(\cos26.6\degree\hat{i}+\sin26.6\degree\hat{j})$

Net force on  $-pq$ is zero, $\bold{F}_1+\bold{F}_2+\bold{F}_3=0$

${\frac{kmpq^2}{(\sqrt{5}a)^2}(\cos26.6\degree\hat{i}-\sin26.6\degree\hat{j})+\frac{kmpq^2}{(\sqrt{5}a)^2}(\cos26.6\degree\hat{i}+\sin26.6\degree\hat{j})-\frac{knpq^2}{(2a)^2}\hat{i}=0}$

$\frac{2kmpq^2}{(\sqrt{5}a)^2}(\cos26.6\degree\hat{i})-\frac{knpq^2}{(2a)^2}\hat{i}=0$

$\frac{2m\cos26.6\degree}{5}-\frac{n}{4}=0$

$m=\frac{5n}{8\cos26.6\degree}=6.3$

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(c)$n=9,m=19$

Net force on $-pq$ is zero.

$\bold{F}_3=\frac{knpq^2}{(x)^2}(-\hat{i})$ , $\bold{F}_1=\frac{kmpq^2}{(\sqrt{a^2+x^2})^2}(\cos\theta\hat{i}-\sin\theta\hat{j})$ ,$\bold{F}_2=\frac{kmpq^2}{(\sqrt{a^2+x^2})^2}(\cos\theta\hat{i}+\sin\theta\hat{j})$

$\bold{F}_1+\bold{F}_2+\bold{F}_3=0$

$\frac{kmpq^2}{(\sqrt{a^2+x^2})^2}(\cos\theta\hat{i}-\sin\theta\hat{j})+\frac{kmpq^2}{(\sqrt{a^2+x^2})^2}(\cos\theta\hat{i}-\sin\theta\hat{j})-\frac{knpq^2}{x^2}\hat{i}=0$

$\frac{2kmpq^2}{(a^2+x^2)}(\cos\theta\hat{i})-\frac{knpq^2}{x^2}\hat{i}=0$

$\frac{2m}{(a^2+x^2)}(\cos\theta)-\frac{n}{x^2}=0$

Since $\cos\theta=\frac{x}{\sqrt{x^2+a^2}}$

$\frac{2mx}{(a^2+x^2)^{3/2}}=\frac{n}{x^2}$

$\frac{2m}{n}=\frac{(a^2+x^2)^{3/2}}{x^3}=\left(1+\frac{a^2}{x^2} \right )^{3/2}$

$\frac{a^2}{x^2}=\left(\frac{2m}{n} \right )^{2/3}-1$

$x=\left[\frac{a^2}{\left(\frac{2m}{n} \right )^{2/3}-1} \right ]^{1/2}$

$x=\frac{a}{\left[ \left(\frac{2m}{n} \right )^{2/3}-1\right ]^{1/2}}=\frac{a}{\left[ \left(\frac{2*19}{9} \right )^{2/3}-1\right ]^{1/2}}=0.62022\,a$

(d)

$x=\frac{a}{\left[ \left(\frac{2m}{n} \right )^{2/3}-1\right ]^{1/2}}$

When $m>>n$, $x\to 0$

And as $\frac{m}{n}\to\frac{1}{2}$ , $x\to+\infty$

Hence range of $\frac{m}{n}$ is $\frac{1}{2}<\frac{m}{n}<&plus;\infty$