As shown in the figure,positive and negative | vedclass

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

Question

(D) The potential at point $A$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:$V_A = \frac{k(1 	imes 10^{-6})}{2} + \frac{k(-1 	imes 10^{-6})}{3}

Vedclass · Accepted Answer

$3000$

Vedclass · Answer

(D) The potential at point $A$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:$V_A = \frac{k(1 	imes 10^{-6})}{2} + \frac{k(-1 	imes 10^{-6})}{3} = k 	imes 10^{-6} \left( \frac{1}{2} - \frac{1}{3} ight) = k 	imes 10^{-6} \left( \frac{1}{6} ight)$$V_A = \frac{9 	imes 10^9 	imes 10^{-6}}{6} = 1.5 	imes 10^3 \ V$The potential at point $B$ due to the charges $+1\ \mu C$ and $-1\ \mu C$ is:$V_B = \frac{k(1 	imes 10^{-6})}{3} + \frac{k(-1 	imes 10^{-6})}{2} = k 	imes 10^{-6} \left( \frac{1}{3} - \frac{1}{2} ight) = k 	imes 10^{-6} \left( -\frac{1}{6} ight)$$V_B = -\frac{9 	imes 10^9 	imes 10^{-6}}{6} = -1.5 	imes 10^3 \ V$The potential difference between $A$ and $B$ is:$V_A - V_B = 1.5 	imes 10^3 - (-1.5 	imes 10^3) = 3.0 	imes 10^3 \ V = 3000 \ V$

As shown in the figure,positive and negative charges of $1\ \mu C$ are placed at two points. Find the potential difference between $A$ and $B$ in $volt$.

Similar Questions

$A$ point charge $q$ of mass $m$ is placed at the center of a ring of radius $R$ carrying a charge $Q$. When it is slightly displaced,the point charge accelerates to infinity along the $x$-axis. The ultimate speed of the point charge is .......

Write an equation for the potential at a point in a uniformly charged spherical shell.

Two spheres $A$ and $B$ of radius $a$ and $b$ respectively are at the same electric potential. The ratio of the surface charge densities of $A$ and $B$ is

The radius of a soap bubble whose potential is $16\,V$ is doubled. The new potential of the bubble will be.....$V$

The electric potential on the surface of a hollow sphere of radius $5 \ cm$ is $10 \ V$. What is the electric potential at its center?

Vedclass Test Series

Exam Paper Generator

Online Exam Module