Question 879: Electrostats & Capacitance

Question

A point charge $$\mathrm{Q}$$ is placed at the centre of the line joining two equal point charges $$+\mathrm{q}$$ and $$+\mathrm{q}$$. The value of $$Q$$ if the system of the charges is in equilibrium, is

Accepted Answer

The forces on charge Q due to the other two charges will be equal and opposite and hence it will be in equilibrium.
Force on $$+\mathrm{q}$$ due to the other two charges should also be equal and opposite. Their magnitudes will be equal if
$$\begin{aligned} & \frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{\mathrm{r}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{qQ}}{\left(\frac{\mathrm{r}}{2}ight)^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \mathrm{qQ}}{\mathrm{r}^2} \ & 	herefore \mathrm{q}=4 \mathrm{Q} \quad 	ext { or } \mathrm{Q}=\frac{\mathrm{q}}{4}\end{aligned}$$
$$	herefore$$ q = 4Q
or Q = $$\frac{q}{4}$$
Q and q should have opposite signs so that the two forces are opposite in direction.
$$	herefore \mathrm{Q=-\frac{q}{4}}$$

Answer

$$\frac{-q}{2}$$

Answer

$$-\frac{9}{4}$$

Answer

$$\frac{+q}{4}$$

Answer

$$\frac{+q}{2}$$

Electrostats & Capacitance

A point charge $\mathrm{Q} is placed at the centre of the line joining two equal point charges +\mathrm{q} and +\mathrm{q}. The value of Q$ if the system of the charges is in equilibrium, is

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