Two metal spheres,one of radius R and the oth | vedclass

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

Question

(D) Initial charges on the spheres are $Q_1 = \sigma (4 \pi R^2)$ and $Q_2 = \sigma (4 \pi (2R)^2) = 16 \pi R^2 \sigma$.Total charge $Q_{total} = Q_1

Vedclass · Accepted Answer

$\sigma_{1}=\frac{5}{3} \sigma, \sigma_{2}=\frac{5}{6} \sigma$

Vedclass · Answer

(D) Initial charges on the spheres are $Q_1 = \sigma (4 \pi R^2)$ and $Q_2 = \sigma (4 \pi (2R)^2) = 16 \pi R^2 \sigma$.Total charge $Q_{total} = Q_1 + Q_2 = 4 \pi R^2 \sigma + 16 \pi R^2 \sigma = 20 \pi R^2 \sigma$.When the spheres are brought into contact,the total charge is redistributed such that they reach the same potential $V = \frac{k Q_1'}{R} = \frac{k Q_2'}{2R}$.This implies $Q_2' = 2 Q_1'$.Since $Q_1' + Q_2' = 20 \pi R^2 \sigma$,we have $Q_1' + 2 Q_1' = 20 \pi R^2 \sigma$,which gives $3 Q_1' = 20 \pi R^2 \sigma$,so $Q_1' = \frac{20}{3} \pi R^2 \sigma$.Then $Q_2' = 2 	imes \frac{20}{3} \pi R^2 \sigma = \frac{40}{3} \pi R^2 \sigma$.The new surface charge densities are $\sigma_1' = \frac{Q_1'}{4 \pi R^2} = \frac{20/3 \pi R^2 \sigma}{4 \pi R^2} = \frac{5}{3} \sigma$.And $\sigma_2' = \frac{Q_2'}{4 \pi (2R)^2} = \frac{40/3 \pi R^2 \sigma}{16 \pi R^2} = \frac{40}{48} \sigma = \frac{5}{6} \sigma$.

Two metal spheres,one of radius $R$ and the other of radius $2R$,respectively,have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them?

Similar Questions

Explain electrostatic shielding with a necessary diagram.

Show that electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface.

Vedclass Test Series

Exam Paper Generator

Online Exam Module