A solid sphere of radius R has charge -q- uniformly distributed over its volume- The distance from its surface which the electrostatic potential is equal to half of the potential the centre is -R2Rdisplaystyle dfrac-R-3-displaystyle dfrac-R-2-

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Toppr Admin · Accepted Answer

Electric potential inside the sphere V- kq12R-3-x2212-r2R2-V- kq32R at the centerPotential out side the sphere is V1-Kq-rnow V1-Vgives r-4R-3so distance from surface 4R-3-R-R-3

A solid sphere of radius $R$ has charge $^{'} q^{'}$ uniformly distributed over its volume. The distance from its surface at which the electrostatic potential is equal to half of the potential at the centre is :
$R$
$2 R$
$\frac{R}{3}$
$\frac{R}{2}$

Electric potential inside the sphere $V =$ $k q \frac{1}{2 R} (3 - \frac{r^{2}}{R^{2}})$
$V =$ $k q \frac{3}{2 R}$ at the center
Potential out side the sphere is $V_{1} = K q / r$
now $V_{1} = V$
gives $r = 4 R / 3$
so distance from surface 4R/3-R=R/3

A solid sphere of radius R has charge ′q′ uniformly distributed over its volume. The distance from its surface at which the electrostatic potential is equal to half of the potential at the centre is :R2RR3R2

Electric potential inside the sphere V= kq12R(3−r2R2)V= kq32R at the centerPotential out side the sphere is V1=Kq/rnow V1=Vgives r=4R/3so distance from surface 4R/3-R=R/3

A solid sphere of radius $R$ has charge $^{'} q^{'}$ uniformly distributed over its volume. The distance from its surface at which the electrostatic potential is equal to half of the potential at the centre is :
$R$
$2 R$
$\frac{R}{3}$
$\frac{R}{2}$

Electric potential inside the sphere $V =$ $k q \frac{1}{2 R} (3 - \frac{r^{2}}{R^{2}})$
$V =$ $k q \frac{3}{2 R}$ at the center
Potential out side the sphere is $V_{1} = K q / r$
now $V_{1} = V$
gives $r = 4 R / 3$
so distance from surface 4R/3-R=R/3