Question 815: Electrostats & Capacitance

Question

Two point charges +10 q and -4 q are located at $\mathrm{x}=0$ and $\mathrm{x}=\mathrm{L}$ respectively. What is the location of a point on the $x$-axis from the origin, which the net electric field due to these two point charges is zero?( $r=$ required distance$)$

Accepted Answer

Let ' $Q$ ' be the point at a distance $r$ from $-4 q$ and $(L+r)$ from 10q. wherein net electric field is zero.
$$\begin{aligned}
& E_1=E_2 \
& \frac{K(10 q)}{(L+r)^2}=\frac{K(4 q)}{r^2} \
& \frac{\sqrt{10}}{L+r}=\frac{2}{r} \
& \frac{\sqrt{5} 	imes \sqrt{2}}{L+r}=\frac{\sqrt{2} 	imes \sqrt{2}}{r}=\frac{2}{r} \
& \sqrt{5} 	imes \sqrt{2} 	imes r=2(L+r) \
& \sqrt{5} 	imes \sqrt{2} 	imes r=\sqrt{2} 	imes \sqrt{2}(L+r) \
& \sqrt{5} r=\sqrt{2} L+\sqrt{2} r \
& \sqrt{5}-\sqrt{2} r=\sqrt{2} L \
& r=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}
\end{aligned}$$
If $B$ is position of charge $-4 q$ then point $Q$ is $r=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ to the right of $B$

Answer

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ right to pt. B

Answer

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ left to pt.A

Answer

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ right to pt.B

Answer

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ left to pt.A

Electrostats & Capacitance

Two point charges +10 q and -4 q are located at \mathrm{x}=0 and \mathrm{x}=\mathrm{L} respectively. What is the location of a point on the x-axis from the origin, which the net electric field due to these two point charges is zero?( r= required distance)

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