A small conducting sphere of radius r is lyin | vedclass

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

Question

(A) The potential $V_{1}$ of the smaller sphere (radius $r$,charge $q$) is due to its own charge and the potential due to the outer sphere (radius $R$

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{r}-\frac{q}{R}\right)$

Vedclass · Answer

(A) The potential $V_{1}$ of the smaller sphere (radius $r$,charge $q$) is due to its own charge and the potential due to the outer sphere (radius $R$,charge $Q$).$V_{1} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}$The potential $V_{2}$ of the bigger sphere (radius $R$,charge $Q$) is due to its own charge and the potential due to the inner sphere (which acts as a point charge at the center for points outside it).$V_{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$The potential difference between the spheres is $V = V_{1} - V_{2}$.$V = \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} \right) - \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R} \right)$$V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$$V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{r} - \frac{q}{R} \right)$

$A$ small conducting sphere of radius $r$ is lying concentrically inside a bigger hollow conducting sphere of radius $R$. The bigger and smaller spheres are charged with $Q$ and $q$ respectively $(Q > q)$ and are insulated from each other. The potential difference between the spheres will be

Similar Questions

The charge given to a hollow sphere of radius $10\, cm$ is $3.2 \times 10^{-19}\, C$. At a distance of $4\, cm$ from its centre,the electric potential will be

Two point charges are located on the $x$-axis. $q_1 = -1 \ \mu C$ at $x = 0$ and $q_2 = +1 \ \mu C$ at $x = 1 \ m$. Find the work done in bringing a third charge $q_3 = +1 \ \mu C$ from infinity to $x = 2 \ m$.

Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively,which are a distance $2L$ apart. $C$ is the midpoint between $A$ and $B$. The work done in moving a charge $+Q$ along the semicircle $CRD$ is

Draw a graph showing the variation of electric potential $V$ with distance $r$ from the center for a uniformly charged spherical shell of radius $R$.

When a charge of $20 C$ is taken from one point to another separated by a distance of $0.2 m$,work of $2 J$ is required to be done. What is the potential difference between the two points?

Vedclass Test Series

Exam Paper Generator

Online Exam Module