In the given figure distance of the point fro | vedclass

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

Question

(c) Suppose electric field is zero at $N$. Hence $|E_A| = |E_B|$
Which gives ${x_1} = \frac{x}{{\sqrt {\frac{{{Q_2}}}{{{Q_1}}}} + 1}} = \frac{{80}}{{

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(c) Suppose electric field is zero at $N$. Hence $|E_A| = |E_B|$
Which gives ${x_1} = \frac{x}{{\sqrt {\frac{{{Q_2}}}{{{Q_1}}}} + 1}} = \frac{{80}}{{\sqrt {\frac{{20}}{{10}}} + 1}} = 33\,cm$

In the given figure distance of the point from $A$ where the electric field is zero is......$cm$

Similar Questions

A charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be.......$N/C$ $(g = 10\,m{s^{ - 1}})$

Time period of a block suspended from the upper plate of a parallel plate capacitor by a spring of stiffness $k$ is $T$. When block is uncharged. If a charge $q$ is given to the block them, the new time period of oscillation will be

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 :

What will be the magnitude of electric field at point $O$ as shown in figure ? Each side of the figure is $I$ and perpendicular to each other.

The unit of intensity of electric field is