Electric potential at a point P due to a poin | vedclass

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

Question

(C) The formula for electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.Giv

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The formula for electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$.Given: $V = 50 \; V$,$Q = 5 	imes 10^{-9} \; C$,and $k = \frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 \; N m^2 C^{-2}$.Substituting the values into the formula: $50 = \frac{9 	imes 10^9 	imes 5 	imes 10^{-9}}{r}$.Simplifying the numerator: $50 = \frac{45}{r}$.Solving for $r$: $r = \frac{45}{50} = 0.9 \; m$.Converting meters to centimeters: $r = 0.9 	imes 100 \; cm = 90 \; cm$.

Electric potential at a point $P$ due to a point charge of $5 \times 10^{-9} \; C$ is $50 \; V$. The distance of $P$ from the point charge is ......... $cm$. (Assume,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \; N m^2 C^{-2}$)

Similar Questions

The electric potentials at two points $P$ and $Q$ are $10 \ V$ and $-4 \ V$ respectively. The work done in moving $100$ electrons from $P$ to $Q$ is:

The value of the electric potential at a point due to a point charge is:

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

Three concentric metallic shells $A, B$ and $C$ of radii $a, b$ and $c$ $(a < b < c)$ have surface charge densities $+\sigma, -\sigma$ and $+\sigma$ respectively. The potential of shell $B$ is

Two particles with charges $+q$ and $-q$ are separated by a distance of $2d$. What is the electric potential at the midpoint of the line joining them?

Vedclass Test Series

Exam Paper Generator

Online Exam Module