Two equal point charges are fixed at x = -a a | vedclass

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

Question

Potential energy $=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ r }$

$|\Delta P |=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ xdx }-\frac{1}{4

Vedclass · Accepted Answer

$x^2$

Vedclass · Answer

Potential energy $=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ r }$

$|\Delta P |=\frac{1}{4 \pi \epsilon_0} q \cdot \frac{ Q }{ xdx }-\frac{1}{4 \pi \epsilon_0} \frac{ qQ }{ x +d x }$

$\lim _i x \rightarrow 0=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q \cdot 2 x}{x^2-(d x)^2}$

Where $dx =0$

so the answer is $x ^2$

Two equal point charges are fixed at $x = -a$ and $x = + \,a$ on the $x$-axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$ ehen it is displaced by a small distance $x$ along the $x$ -axis is apporximately proportional to

Similar Questions

Two point charges $+8q$ and $-2q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the $x-$ axis at which net electric field due to these two point charges is zero, is

The charge $q$ on a capacitor varies with voltage as shown in figure. The area of the triangle $AOB$ is proportional to

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

The potential $V$ is varying with $x$ and $y$ as $V = \frac{1}{2}({y^2} - 4x)\,volts$ The field at $(1\,m,\,1\,m)$ is