Two charges 9e and 3e are placed at a distanc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the electric field be zero at a distance $x_1$ from the charge $Q_1 = 9e$ and $x_2$ from the charge $Q_2 = 3e$,where $x_1 + x_2 = r$.At the nu

Vedclass · Accepted Answer

At a distance of $\frac{r}{1 + \sqrt{1/3}}$ from the $9e$ charge.

Vedclass · Answer

(A) Let the electric field be zero at a distance $x_1$ from the charge $Q_1 = 9e$ and $x_2$ from the charge $Q_2 = 3e$,where $x_1 + x_2 = r$.At the null point,the magnitudes of the electric fields produced by both charges must be equal:$\frac{k Q_1}{x_1^2} = \frac{k Q_2}{x_2^2}$$\frac{9e}{x_1^2} = \frac{3e}{x_2^2}$$\frac{x_1^2}{x_2^2} = \frac{9e}{3e} = 3$$\frac{x_1}{x_2} = \sqrt{3} \implies x_1 = x_2 \sqrt{3}$Since $x_1 + x_2 = r$,we have $x_2 \sqrt{3} + x_2 = r \implies x_2(1 + \sqrt{3}) = r \implies x_2 = \frac{r}{1 + \sqrt{3}}$ (distance from $3e$ charge).For $x_1$,$x_1 = r - x_2 = r - \frac{r}{1 + \sqrt{3}} = r \left( \frac{1 + \sqrt{3} - 1}{1 + \sqrt{3}} \right) = \frac{r \sqrt{3}}{1 + \sqrt{3}} = \frac{r}{1 + \frac{1}{\sqrt{3}}}$.Thus,the electric field is zero at a distance of $\frac{r}{1 + \sqrt{1/3}}$ from the $9e$ charge.

Two charges $9e$ and $3e$ are placed at a distance $r$. At what distance will the electric field be zero?

Similar Questions

In the given figure,the electric field at the center $O$ due to section $AB$ of a uniformly charged ring is $\overrightarrow{E}$. What will be the electric field at $O$ due to the remaining section $ACB$?

The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\,N/C$ just above its surface is:

The number of electrons to be put on a spherical conductor of radius $0.1 \ m$ to produce an electric field of $0.036 \ N/C$ just above its surface is $...... \times 10^4$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module