A spherical metal shell A of radius R_A and a | vedclass

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

Question

(B) When two conductors are connected by a wire,charge flows until their potentials become equal. Let the final charges be $Q_A$ and $Q_B$.Since $V_A

Vedclass · Accepted Answer

$(A, B, C, D)$

Vedclass · Answer

(B) When two conductors are connected by a wire,charge flows until their potentials become equal. Let the final charges be $Q_A$ and $Q_B$.Since $V_A = V_B$,we have $\frac{kQ_A}{R_A} = \frac{kQ_B}{R_B}$,which implies $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$.Since $R_A > R_B$,it follows that $Q_A > Q_B$. Thus,$(B)$ is correct.For a spherical shell,the electric field inside is always zero due to Gauss's Law. Thus,$(A)$ is correct.The surface charge density is $\sigma = \frac{Q}{4\pi R^2}$. Therefore,$\frac{\sigma_A}{\sigma_B} = \frac{Q_A}{4\pi R_A^2} \cdot \frac{4\pi R_B^2}{Q_B} = \frac{Q_A}{Q_B} \cdot \frac{R_B^2}{R_A^2} = \frac{R_A}{R_B} \cdot \frac{R_B^2}{R_A^2} = \frac{R_B}{R_A}$. Thus,$(C)$ is correct.The electric field on the surface is $E = \frac{\sigma}{\epsilon_0}$. Since $\frac{\sigma_A}{\sigma_B} = \frac{R_B}{R_A} Therefore,all options $(A, B, C, D)$ are correct.

Similar Questions

The metal plate on the left in the figure has a charge of $+q$. The metal plate on the right has a charge of $-2q$. What charge will flow to the earth when switch $S$ is closed, given that the middle plate is initially neutral?

Two metallic spheres of radii $a$ and $b$ are placed far away from each other and are connected by a thin conducting wire. The total charge on them is $Q$. Calculate the potential of each sphere.

Four very large metal plates are given the charges as shown in the figure. The middle two plates are then connected through a wire. Find the charge that will flow through the wire.

$A$ metallic spherical shell has an inner radius $R_1$ and an outer radius $R_2$. $A$ charge $Q$ is placed at the center of the spherical cavity. What will be the surface charge density on $(i)$ the inner surface,and $(ii)$ the outer surface?

As shown in the figure,a point charge $Q$ is placed at the centre of a conducting spherical shell of inner radius $a$ and outer radius $b$. The electric field due to charge $Q$ in three different regions $I$,$II$,and $III$ is given by: $(I: r < a, II: a < r < b, III: r > b)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module