Three concentric metallic spherical shells A, | vedclass

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

Question

(C) The charges on the shells are given by $q = 	ext{surface charge density} 	imes 	ext{area}$.For shell $A$ (radius $a$): $q_A = -\sigma(4\pi a^2)

Vedclass · Accepted Answer

$(\sigma/\varepsilon_0) [b - a - c]$

Vedclass · Answer

(C) The charges on the shells are given by $q = 	ext{surface charge density} 	imes 	ext{area}$.For shell $A$ (radius $a$): $q_A = -\sigma(4\pi a^2)$.For shell $B$ (radius $b$): $q_B = +\sigma(4\pi b^2)$.For shell $C$ (radius $c$): $q_C = -\sigma(4\pi c^2)$.The potential at the surface of shell $A$ is the sum of potentials due to all three shells:$V_A = V_{A,A} + V_{A,B} + V_{A,C} = \frac{1}{4\pi\varepsilon_0} [\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}]$.Substituting the values:$V_A = \frac{1}{4\pi\varepsilon_0} [\frac{-\sigma(4\pi a^2)}{a} + \frac{\sigma(4\pi b^2)}{b} + \frac{-\sigma(4\pi c^2)}{c}]$.$V_A = \frac{1}{4\pi\varepsilon_0} [-\sigma(4\pi a) + \sigma(4\pi b) - \sigma(4\pi c)]$.$V_A = \frac{\sigma}{\varepsilon_0} [-a + b - c] = \frac{\sigma}{\varepsilon_0} [b - a - c]$.

Three concentric metallic spherical 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 $A$ is:

Similar Questions

If $4 \times 10^{20} \text{ eV}$ of energy is required to move a charge of $0.25 \text{ C}$ between two points,what will be the potential difference between them in volts?

The electric potential at the center $O$ of two concentric half rings of radii $R_1$ and $R_2$, having the same linear charge density $\lambda$, is $(\epsilon_0 = \text{permittivity of free space})$:

An electric charge $10^{-3} \, \mu C$ is placed at the origin $(0, 0)$ of an $X-Y$ coordinate system. Two points $A$ and $B$ are situated at $(\sqrt{2}, \sqrt{2})$ and $(2, 0)$ respectively. The potential difference between the points $A$ and $B$ will be $V$.

If four charges $+12 \text{ nC}, -20 \text{ nC}, +32 \text{ nC}$ and $-15 \text{ nC}$ are arranged at the four vertices of a square of side $\sqrt{2} \text{ m}$,then the net electric potential at the centre of the square due to these four charges is (in $\text{ V}$)

An electron of charge $e$ coulomb passes through a potential difference of $V$ volts. Its energy in joules will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module