Question 809: Electrostats & Capacitance

Question

Charges $3 \mathrm{Q}, \mathrm{q}$ and Q are placed along x -axis at positions $\mathrm{x}=0, \mathrm{x}=\frac{1}{3}$ and $\mathrm{x}=1$ respectively. When the force on charge Q is zero, the value of $q$ is

Accepted Answer

Force $F_1$ due to charge $3 Q$ on $Q$ will be,
$$\mathrm{F}_1=\frac{1}{4 \pi \varepsilon_0} \frac{3 \mathrm{Q}^2}{l^2}$$
Force $F_2$ due to charge $q$ on charge $Q$ will be,
$$\mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\left(\frac{2 l}{3}ight)^2}$$
$	herefore \quad \mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{9 \mathrm{Qq}}{4 l^2}$
For the force on charge Q to be zero,
$\mathrm{F}_1+\mathrm{F}_2=0$ or $\mathrm{F}_1=-\mathrm{F}_2$
$$\begin{array}{ll}
	herefore & \frac{3 \mathrm{Q}^2}{l^2}=-\frac{9 \mathrm{Qq}}{4 l^2} \
	herefore & \mathrm{Q}=-\frac{3 \mathrm{q}}{4} \
	herefore & \mathrm{q}=-\frac{4}{3} \mathrm{Q}
\end{array}$$

Answer

$\frac{\mathrm{Q}}{3}$

Answer

$-\frac{\mathrm{Q}}{3}$

Answer

$\frac{4}{3} \mathrm{Q}$

Answer

$-\frac{4}{3} \mathrm{Q}$

Electrostats & Capacitance

Charges 3 \mathrm{Q}, \mathrm{q} and Q are placed along x -axis at positions \mathrm{x}=0, \mathrm{x}=\frac{1}{3} and \mathrm{x}=1 respectively. When the force on charge Q is zero, the value of q is

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