Two spheres A and B of radius a and b respect | vedclass

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

Question

(B) Given that the electric potential of both spheres is the same,i.e.,$V_A = V_B$.Using the formula for the potential of a charged sphere $V = \frac{

Vedclass · Accepted Answer

$b/a$

Vedclass · Answer

(B) Given that the electric potential of both spheres is the same,i.e.,$V_A = V_B$.Using the formula for the potential of a charged sphere $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r}$,we have:$\frac{1}{4\pi\epsilon_0} \cdot \frac{Q_A}{a} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_B}{b} \implies \frac{Q_A}{Q_B} = \frac{a}{b}$......$(i)$Surface charge density $\sigma$ is defined as $\sigma = \frac{Q}{4\pi r^2}$.Therefore,the ratio of surface charge densities is:$\frac{\sigma_A}{\sigma_B} = \frac{Q_A}{4\pi a^2} \cdot \frac{4\pi b^2}{Q_B} = \frac{Q_A}{Q_B} \cdot \frac{b^2}{a^2}$Substituting the ratio from equation $(i)$:$\frac{\sigma_A}{\sigma_B} = \frac{a}{b} \cdot \frac{b^2}{a^2} = \frac{b}{a}$.

Two spheres $A$ and $B$ of radius $a$ and $b$ respectively are at the same electric potential. The ratio of the surface charge densities of $A$ and $B$ is

Similar Questions

$A$ conducting sphere of radius $R$ is given a charge $Q$. The electric potential and the electric field at the centre of the sphere respectively are

The dimensions of electric potential are

If an electron moves from rest from a point at which potential is $50\, V$ to another point at which potential is $70\, V$,then its kinetic energy in the final state will be:

For a conducting sphere of radius $R$ carrying a charge $Q$,what is the electric potential at a distance $x$ from the center inside the sphere?

The maximum electric field that can be held in air without producing ionization of air is $10^7\,V/m$. The maximum potential,therefore,to which a conducting sphere of radius $0.10\,m$ can be charged in air is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module