Tag: electrostatics

Questions Related to electrostatics

The potential in certain region is given as $V = 2x^2$, then the charge density of that region is 

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $-\dfrac{4x}{\varepsilon _0}$

  2. $-\dfrac{4}{\varepsilon _0}$

  3. $-4 \varepsilon _0$

  4. $-2 \varepsilon _0$

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Correct Option: C
Explanation:

By Gauss's law, $\nabla^2V=-\frac{\rho}{\varepsilon _0}$

So, $\frac{\partial^2V}{\partial x^2}=-\frac{\rho}{\varepsilon _0} ...(1)$
Give, $V=2x^2$
or $\frac{\partial V}{\partial x}=4x$
or $\frac{\partial^2V}{\partial x^2}=4$
Now from (1), $\rho=-4\varepsilon _0$

A particle A has charge +q and a particle B has a charge +9q with each of them having the same mass m.If both the particles are allowed to all from rest through the same potential difference, then the ratio of their speed is

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $1:2$

  2. $1:\sqrt 3$

  3. $1:2\sqrt { 2 } $

  4. none of these

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Correct Option: B
Explanation:
A charge $=+q$
B charge $=+9q$
mass $=m$
same difference $=?$
ratio of their speed $=?$
$V=\sqrt { mgh } $
$V=$
now, the working formula is,
$\dfrac { { K } _{ a } }{ { K } _{ b } } =\dfrac { q\times V }{ 9q\times V } $
$\dfrac { -\dfrac { 1 }{ 2 } { mV } _{ a }^{ 2 } }{ \dfrac { 1 }{ 2 } { mV } _{ 0 }^{ 2 } } =\dfrac { 1 }{ 9 } $
or,  $\dfrac { { V } _{ a } }{ { V } _{ b } } =\dfrac { 1 }{ 3 } $
It is the ratio between the kinetic energies between the two parties.
$\dfrac { 1 }{ 3 } =\dfrac { { V } _{ A }^{ 2 } }{ { V } _{ b }^{ 2 } } $
or,  $\dfrac { { V } _{ A } }{ { V } _{ b } } =\dfrac { 1 }{ \sqrt { 3 }  } $
or,  ${ V } _{ A }:{ V } _{ b }=1:\sqrt { 3 } $

Positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point A at distance r(r> R) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion.

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $\displaystyle \left[\frac{1}{2 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$

  2. $\displaystyle \left[\frac{1}{2 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{8}\right) \right]^{1/2}$

  3. $\displaystyle \left[\frac{1}{2 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} - \frac{3}{8}\right) \right]^{1/2}$

  4. $\displaystyle \left[\frac{1}{4 \pi \varepsilon _0} \frac{Qq}{Rm} \left(\frac{r-R}{r} + \frac{3}{4}\right) \right]^{1/2}$

Show answer
Correct Option: B
Explanation:

Initial energy are kinetic and potential energy given by-


$K _i=\dfrac{1}{2}mv^2$       and $U _i=\dfrac{Qq}{4\pi\epsilon _o r}$

Finally at last point , its velocity get reduced to $0$, and potential at a point inside sphere is given by-

$V=\dfrac{Q}{4\pi\epsilon _o}\dfrac{3R^2-r^2}{2R^3}$

At $r=\dfrac{R}{2}$, $U _f=qV$

$\implies U _f=\dfrac{Qq}{4\pi\epsilon _o}\dfrac{3R^2-\dfrac{R^2}{4}}{2R^3}$

$\implies U _f=\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon _oR}$

Now applying conservation of mechanical energy-

$K _i+U _i=K _f+U _f$

$\implies \dfrac{1}{2}mv^2+\dfrac{Qq}{4\pi\epsilon _o r}=0+\dfrac{11}{8}\dfrac{Qq}{4\pi\epsilon _o R}$

$\implies mv^2=\dfrac{Qq}{2\pi\epsilon _o}\left(\dfrac{11}{8R}-\dfrac{1}{r}\right)$

$\implies v^2=\dfrac{Qq}{2\pi\epsilon _oRm}\left(\dfrac{11}{8}-\dfrac{R}{r}\right)$

$\implies v^2=\dfrac{Qq}{2\pi\epsilon _oRm}\left(1+\dfrac{3}{8}-\dfrac{R}{r}\right)$

$\implies v^2=\dfrac{1}{2\pi\epsilon _o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)$

$\implies v=\sqrt{\dfrac{1}{2\pi\epsilon _o}\dfrac{Qq}{Rm}\left(\dfrac{r-R}{r}+\dfrac{3}{8}\right)}$

Answer-(B)

A particle of mass $10^{-3}kg$ and charge $5\mu C$ is thrown at a speed $20\ m\ s^{-1}$ against a uniform electric field of strength $2\times 10^{5}N\ C^{-1}$. How much distance will it travel before coming to rest momentarily?

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $0.1\ m$

  2. $0.3\ m$

  3. $0.05\ m$

  4. $0.2\ m$

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Correct Option: D
Explanation:

Force on particle$=F=qE$ in opposite direction of motion


And, $F=ma=qE$

$\implies a=\dfrac{qE}{m}=\dfrac{5\times 10^{-6}\times 2\times 10^5}{10^{-3}}$

$\implies a=10^3m/s^2$ and this acceleration is negative since particle is thrown against force.
And final velocity is $v=0$

Using  $v^2-u^2=2as$

$\implies 0-20^2=-2\times 1000\times s$

$\implies s=0.2m$

Answer-(D)

The electric potential energy of a uniformly charged thin spherical shell of radius 'R' having a total charge 'Q' is

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $\dfrac{KQ^2}{4R}$

  2. $\dfrac{KQ^2}{6R}$

  3. $\dfrac{KQ^2}{8R}$

  4. $\dfrac{KQ^2}{16R}$

Show answer
Correct Option: C
Explanation:
The electric potential energy of a uniform charged.
radius $=R$
charge $=Q$
electric potential
$E=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } .\dfrac { { Q }^{ 2 } }{ R }  \right) $
   $=\dfrac { 1 }{ 8 } K\dfrac { { Q }^{ 2 } }{ R } $      ($\because$   $\dfrac { 1 }{ \pi { \epsilon  } _{ 0 } } =K$)

A uniform electric field of magnitude $290 V/m$ is directed in the positive $x$ direction. A $+13.0 \mu C$ charge moves from the origin to the point $(x, y) = (20.0 cm, 50.0 cm).$

What is the change in the potential energy of the charge field system?

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $-754J$

  2. $-754mJ$

  3. $-754kJ$

  4. $-754\mu J$

Show answer
Correct Option: D
Explanation:
Given electric field, $\overrightarrow{E}=290\, V/m$ directed along $+x$ direction.
Charge $=+13\mu C$ moves from origin to point $(x,y)=(20\, cm , 50\, cm)$.
We have to find the charge in potential energy of the charge field system.
We know, the change in potential energy  $=$ charge $\times $ change in potential
i.e, $\Delta U=q\Delta V$
But, $\Delta V=-Ed$
$\Delta U=-qEd$
$=-13\times 10^{-6}\times 290\times 20\times 10^{-2}$
$=0.000754\, J$
$=-754\times 10^{-6}\, J$
Here $d=20\, cm$, since electric field is along positive $x-$axis.