Question 792: Electrostats & Capacitance

Question

Two charges $\mathrm{q}_1=+6_{\mathrm{q}}$ and $\mathrm{q}_2=-3 \mathrm{q}$ placed as shown in figure. A proton is placed on x -axis away from $\mathrm{q}_2$. To remain proton in equilibrium, the distance between $\mathrm{q}_1$ and proton is

Accepted Answer

Let ' $a$ ' be the distance between $q_2$ and $p$
From figure, Distance between $\mathrm{q}_1$ and $p=L+a$
For proton to be in equilibrium, it should be placed at point where electric field due to $q_1$ and $q_2$ is zero.
$$\begin{array}{ll} 
& E_1+E_2=0 \
	herefore & \frac{k q_1}{r_1^2}+\frac{k q_2}{r_2^2}=0 \
	herefore & \frac{+6 q}{(L+a)^2}+\frac{(-3 q)}{a^2}=0 \
	herefore & \frac{6 q}{(L+a)^2}=\frac{3 q}{a^2} \
	herefore & \frac{2}{(L+a)^2}=\frac{1}{a^2} \
	herefore & \frac{\sqrt{2}}{L+a}=\frac{1}{a} \quad 	ext{...... (Taking sq. root)}\
	herefore & \sqrt{2} a=L+a \
\Rightarrow & \sqrt{2} a-a=L \
\Rightarrow & a=\frac{L}{\sqrt{2}-1}
\end{array}$$
$$\begin{aligned}
&	herefore \quad 	ext { Distance between } \mathrm{q}_1 	ext { and } \mathrm{p}=\mathrm{L}+\mathrm{a}\
&\begin{aligned}
& =L+\frac{L}{\sqrt{2}-1}=\frac{(\sqrt{2}-1) L+L}{\sqrt{2}-1}=\frac{\sqrt{2} L}{\sqrt{2}-1} \
& =\frac{\sqrt{2} L}{\sqrt{2}-1}
\end{aligned}
\end{aligned}$$

Answer

$\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right) \mathrm{L}$

Answer

$\mathrm{2 L}$

Answer

$\frac{\mathrm{L}}{2}$

Answer

$\left(\frac{\sqrt{2}}{\sqrt{2}+1}\right) \mathrm{L}$

Electrostats & Capacitance

Two charges \mathrm{q}_1=+6_{\mathrm{q}} and \mathrm{q}_2=-3 \mathrm{q} placed as shown in figure. A proton is placed on x -axis away from \mathrm{q}_2. To remain proton in equilibrium, the distance between \mathrm{q}_1 and proton is

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