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The electric field at $20 \,cm$ from the centre of a uniformly charged non-conducting sphere of radius $10 \,cm$ is $E$. Then at a distance $5 \,cm$ from the centre it will be
$16 E$
$4 E$
$2 E$
Zero
Solution
(c)
$R=10 \,cm , \quad r=20 \,cm$
$E=\frac{k Q}{(0.2)^2}$
Now at $r=5 \,cm$
$E^{\prime}=\frac{k Q(0.05)}{(0.1)^3}$
Now, $\frac{E^{\prime}}{E}=\frac{(0.05)}{(0.1)^3}(0.2)^2=2$
$E=2 E$
Similar Questions
The electric field $E$ is measured at a point $P (0,0, d )$ generated due to various charge distributions and the dependence of $E$ on $d$ is found to be different for different charge distributions. List-$I$ contains different relations between $E$ and $d$. List-$II$ describes different electric charge distributions, along with their locations. Match the functions in List-$I$ with the related charge distributions in List-$II$.
List-$I$ | List-$II$ |
$E$ is independent of $d$ | A point charge $Q$ at the origin |
$E \propto \frac{1}{d}$ | A small dipole with point charges $Q$ at $(0,0, l)$ and $- Q$ at $(0,0,-l)$. Take $2 l \ll d$. |
$E \propto \frac{1}{d^2}$ | An infinite line charge coincident with the x-axis, with uniform linear charge density $\lambda$ |
$E \propto \frac{1}{d^3}$ | Two infinite wires carrying uniform linear charge density parallel to the $x$-axis. The one along ( $y=0$, $z =l$ ) has a charge density $+\lambda$ and the one along $( y =0, z =-l)$ has a charge density $-\lambda$. Take $2 l \ll d$ |
plane with uniform surface charge density |