Three concentric spherical shells have radii | vedclass

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

Question

(D) Let the charges on the shells be $q_1, q_2,$ and $q_3$ respectively.$q_1 = 4\pi a^2\sigma, q_2 = -4\pi b^2\sigma, q_3 = 4\pi c^2\sigma$.The potent

Vedclass · Accepted Answer

$V_C = V_A 
eq V_B$

Vedclass · Answer

(D) Let the charges on the shells be $q_1, q_2,$ and $q_3$ respectively.$q_1 = 4\pi a^2\sigma, q_2 = -4\pi b^2\sigma, q_3 = 4\pi c^2\sigma$.The potential at the surface of shell $a$ is $V_A = \frac{kq_1}{a} + \frac{kq_2}{b} + \frac{kq_3}{c} = k4\pi\sigma(a - b + c)$.The potential at the surface of shell $b$ is $V_B = \frac{kq_1}{b} + \frac{kq_2}{b} + \frac{kq_3}{c} = k4\pi\sigma(\frac{a^2}{b} - b + c)$.The potential at the surface of shell $c$ is $V_C = \frac{kq_1}{c} + \frac{kq_2}{c} + \frac{kq_3}{c} = \frac{k4\pi\sigma}{c}(a^2 - b^2 + c^2)$.Given $c = a + b$,then $c^2 = (a + b)^2 = a^2 + b^2 + 2ab$,so $a^2 - b^2 + c^2 = a^2 - b^2 + a^2 + b^2 + 2ab = 2a^2 + 2ab = 2a(a + b) = 2ac$.Substituting this into $V_C$: $V_C = \frac{k4\pi\sigma}{c}(2ac) = k8\pi\sigma a$.Also,$V_A = k4\pi\sigma(a - b + a + b) = k4\pi\sigma(2a) = k8\pi\sigma a$.Thus,$V_A = V_C 
eq V_B$.

Three concentric spherical shells have radii $a, b,$ and $c$ $(a < b < c)$ and have surface charge densities $\sigma, -\sigma,$ and $\sigma$ respectively. If the potentials on the surfaces of these shells are $V_A, V_B,$ and $V_C$ respectively,then for $c = a + b$,which of the following is true?

Similar Questions

If $\int \frac{7x^8+8x^7}{(1+x+x^8)^2} dx = f(x) + c$,then $f(x)$ is equal to

The classification of $Monera$ is based on which of the following?

At $298 \ K$,the $emf$ of the cell is .......... $V$.
$Pt | H_{2(2 \ atm)} | H^{+}_{(0.002 \ M)} || H^{+}_{(0.1 \ M)} | H_{2(1 \ atm)} | Pt$

Vedclass Test Series

Exam Paper Generator

Online Exam Module