Two charges q and 3 q are separated by a distance ' r ' in air. At a distance x from charge

English

Gujarati

Hindi

1. Electric Charges and Fields

hard

Two charges $q$ and $3 q$ are separated by a distance ' $r$ ' in air. At a distance $x$ from charge $q$, the resultant electric field is zero. The value of $x$ is :

A

$\frac{(1+\sqrt{3})}{r}$

B

$\frac{\mathrm{r}}{3(1+\sqrt{3})}$

C

$\frac{r}{(1+\sqrt{3})}$

D

$r(1+\sqrt{3})$

(JEE MAIN-2024)

Solution

$\left(\vec{E}_{\text {net }}\right)_p=0$

$\frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} \cdot 3 \mathrm{q}}{(\mathrm{r}-\mathrm{x})^2}$

$(\mathrm{r}-\mathrm{x})^2=3 \mathrm{x}^2$

$\mathrm{r}-\mathrm{x}=\sqrt{3} \mathrm{x}$

$\mathrm{x}=\frac{\mathrm{r}}{\sqrt{3}+1}$

Standard 12

Physics

Share

Similar Questions

A flat circular disc has a charge $ + Q$ uniformly distributed on the disc. A charge $ + q$ is thrown with kinetic energy $E$ towards the disc along its normal axis. The charge $q$ will

easy

Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is

hard

Infinite charges of magnitude $q$ each are lying at $x =1,\, 2,\, 4,\, 8…$ meter on $X$-axis. The value of intensity of electric field at point $x = 0$ due to these charges will be

medium

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,\frac{V}{m}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10\,\frac{m}{{{s^2}}}$)

hard

(JEE MAIN-2019)

An oil drop carries six electronic charges, has a mass of $1.6 \times 10^{-12} g$ and falls with a terminal velocity in air. The magnitude of vertical electrical electric field required to make the drop move upward with the same speed as was formely moving is ……..$kN/C$

medium