A hollow charged metal sphere has radius R. I | vedclass

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

Question

(B) The potential at the surface of the sphere is $V_s = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.The potential at a distance $5R$ from the centre is $V_

Vedclass · Accepted Answer

$\frac{V}{20R}$

Vedclass · Answer

(B) The potential at the surface of the sphere is $V_s = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.The potential at a distance $5R$ from the centre is $V_p = \frac{1}{4\pi\epsilon_0} \frac{q}{5R}$.The potential difference $V$ is given by $V = V_s - V_p = \frac{q}{4\pi\epsilon_0} (\frac{1}{R} - \frac{1}{5R}) = \frac{q}{4\pi\epsilon_0} (\frac{4}{5R}) = \frac{q}{5\pi\epsilon_0 R}$.From this,we can express the charge $q$ as $q = \frac{5\pi\epsilon_0 RV}{1}$.The electric field intensity $E$ at a distance $r = 5R$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{(5R)^2} = \frac{q}{100\pi\epsilon_0 R^2}$.Substituting the value of $q$ in terms of $V$: $E = \frac{1}{100\pi\epsilon_0 R^2} \cdot (5\pi\epsilon_0 RV) = \frac{5}{100} \frac{V}{R} = \frac{V}{20R}$.

$A$ hollow charged metal sphere has radius $R$. If the potential difference between its surface and a point at a distance $5R$ from the centre is $V$,then the magnitude of the electric field intensity at a distance $5R$ from the centre of the sphere is:

Similar Questions

The electric potential inside a charged hollow conducting sphere is . . . . . . .

$A$ hollow metal sphere of radius $5\, cm$ is charged such that the potential on its surface is $10\, V$. The potential at the centre of the sphere is

If $27$ charged water droplets, each of radius $10^{-3} \,m$ and charge $10^{-12} \,C$ coalesce to form a single big spherical drop, then the potential of the big drop is (in $\,V$)

The work done in carrying a charge of $5\,\mu C$ from a point $A$ to a point $B$ in an electric field is $10\,mJ$. The potential difference $(V_B - V_A)$ is then

Six charges $+q, -q, +q, -q, +q$ and $-q$ are fixed at the corners of a hexagon of side $d$ as shown in the figure. The work done in bringing a charge $q_0$ to the centre of the hexagon from infinity is: ($\varepsilon_0$ = permittivity of free space)

Vedclass Test Series

Exam Paper Generator

Online Exam Module