Question 357: Electrostats & Capacitance

Question

A point particle of charge $Q$ is located at $P$ along the axis of an electric dipole 1 at a distance $r$ as shown in the figure. The point P is also on the equatorial plane of a second electric dipole 2 at a distance r. The dipoles are made of opposite charge q separated by a distance $2 a$. For the charge particle at P not to experience any net force, which of the following correctly describes the situation?

Accepted Answer

$\begin{aligned} & \frac{\mathrm{kq}}{(\mathrm{r}-\mathrm{a})^2}=\frac{\mathrm{kq}}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{kq}}{\left(\mathrm{r}^2+\mathrm{a}^2ight)} \cos 	heta \ & \frac{1}{(\mathrm{r}-\mathrm{a})^2}=\frac{1}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2ight)^{\frac{3}{2}}} \ & \frac{1}{(\mathrm{r}-\mathrm{a})^2}-\frac{1}{(\mathrm{r}+\mathrm{a})^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2ight)^{\frac{3}{2}}} \ & \frac{4 \mathrm{ra}}{\left(\mathrm{r}^2-\mathrm{a}^2ight)^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2ight)^{\frac{3}{2}}}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{2 r}{\left(r^2-a^2ight)^2}=\frac{1}{\left(r^2+a^2ight)^{\frac{3}{2}}} \ & \frac{4 r^2}{\left(r^2-a^2ight)^4}=\frac{1}{\left(r^2+a^2ight)^3} \ & \Rightarrow 4 r^2\left(r^2+a^2ight)^3=\left(r^2-a^2ight)^4 \ & 4 r^8\left(1+\frac{a^2}{r^2}ight)^3=r^8\left(1-\frac{a^2}{r^2}ight)^4 \ & 4\left(1+\frac{a^2}{r^2}ight)^3=\left(1-\frac{a^2}{r^2}ight)^4\end{aligned}$
Exact value cannot be solved in exam for this equation to be true
$$ \left|\frac{\mathrm{a}}{\mathrm{r}}ight|>1 \Rightarrow \mathrm{a}>\mathrm{r} $$
But point charge $Q$ lies between charges of dipole
1 hence electric field cannot be zero.
There for it should be bonus.
But by solving from mathematical software we are getting $\mathrm{a} / \mathrm{r} \approx 3$.

Answer

$\frac{a}{r} \sim 0.5$

Answer

$\frac{a}{r} \sim 3$

Answer

$\frac{a}{r} \sim 10$

Answer

$\frac{a}{r} \sim 20$

Electrostats & Capacitance

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