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Potential energy of various configurations - class-XII

Description: potential energy of various configurations
Number of Questions: 29
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Tags: electrostatic potential and capacitance physics electrostatics
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Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding . The charges on the two rings are + q and -q . The potential difference between the centres of the two rings is

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

  1. $\frac {{ Q.R }\quad} {\quad { 4\pi }{ \varepsilon } _{ 0 }{ d }^{ 2 }}$

  2. $\frac { Q }{ 2\pi { \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { R }^{ 2 }+{ d }^{ 2 } } } ]$

  3. $\frac { Q }{ { 4\pi \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { { R }^{ 2 } }+{ { d }^{ 2 } } } } ]$

  4. 0

Show answer
Correct Option: A
Explanation:

Two thin wire ring each having a radius $=R$

distance $=d$
two rings are $+q$ and $-q$.
$PD=\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \dfrac { QR }{ { d }^{ 2 } } $
Now in this question, we get the solution.

what is the potential difference between two points, if 2J of work must be done to move a 4 mC charge from one point to another is:

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

  1. 50 V

  2. 500 V

  3. 5 V

  4. 5000 V

Show answer
Correct Option: B
Explanation:

Answer is B.

The total work done = energy transferred.
so, we might see the equation energy = voltage x charge, E = V * Q, written as, 
work = voltage x charge, W = V * Q.
In this case, the charge is 4 mC, that is, 0.004 C and work done is 2 J.
Therefore, V=W/Q = 2/0.004 = 500 V.
Hence, the potential difference between two points if 2 J of work must be done to move a 4 mC charge from one point to another is 500 V.

A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is_______

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

  1. zero

  2. positive

  3. negative

  4. zero if the charge Q is at the center and nonzero otherwise.

Show answer
Correct Option: A
Explanation:

The net displacement round one complete circle is 0.

So, the work done is 0.

$100J$ of work is done when $2 \mu C$ charge is moved in an electric field between two points. The p.d. between the points is

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

  1. $2\times10^{-4}V$

  2. $2\times10^{-8}V$

  3. $2\times10^{-6}V$

  4. $5\times10^{7}V$

Show answer
Correct Option: D
Explanation:

We know work done $=q\Delta V $
Where $\Delta V $ is change in potential $V _2-V _1$
$100=\Delta V\times 2\times 10^{-6}$
$50\times 10^6=\Delta V$
$\therefore p\cdot d= 5\times 10^7 V$

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

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

  1. zero

  2. $10$ volts

  3. Same as at a point $5\ cm$ away from the surface.

  4. Same as at a point $25\ cm$ away from the surface.

Show answer
Correct Option: B
Explanation:
For hollow spherical conductor -
$\Rightarrow$ Potential on its surface $=$ Potential on center $= 10\;volts.$
Because, the electric field inside the spherical shell is zero, due to which the potential inside the shell is constant and will be equal to the potential at the surface.
Hence, the answer is $10\;volts.$

A spherical shell of radius $R _1$ with uniform charge $q$ is expanded to a radius $R _2$. Find the work performed by the electric forces during the shell expansion from $R _1$ to radius $R _2$.

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

  1. $\dfrac{q^2}{2\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

  2. $\dfrac{q^2}{3\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

  3. $\dfrac{q^2}{5\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

  4. $\dfrac{q^2}{8\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

Show answer
Correct Option: D
Explanation:
Spherical shell of radius $={ R } _{ 1 }$
charge $=q$
work done $=W=q\left( { V } _{ B }-{ V } _{ A } \right) $
${ V } _{ A }=\dfrac { { K } _{ q } }{ { R } _{ 1 } } $
${ V } _{ B }=\dfrac { { K } _{ q } }{ { R } _{ 2 } } $
$W=q\left( \dfrac { { K } _{ q } }{ { R } _{ 1 } } -\dfrac { { K } _{ q } }{ { R } _{ 2 } }  \right) $
     $=\dfrac { { q }^{ 2 } }{ 8\pi { \epsilon  } _{ 0 }{ n } _{ 0 } } \left( \dfrac { 1 }{ { R } _{ 1 } } -\dfrac { 1 }{ { R } _{ 2 } }  \right) $    (Proved).

Two insulated charged spheres of radii ${R} _{1}$ and ${R} _{2}$ having charges ${Q} _{1}$ and ${Q} _{2}$ respectively are connected to each other, then there is:

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

  1. no change in the energy of the sytem

  2. an increase in the energy of the system

  3. always a decrease in the energy of the system

  4. a decrease in energy of the system unless ${q} _{1}{R} _{2}={q} _{2}{R} _{1}$

Show answer
Correct Option: D
Explanation:

Two insulated charged spheres of radii $ R _1,R _2$having charges $q _1,q _2$respectively are brought in contact with each other.

Charges will flow across the point of contact from where they are connected until their potential at their surfaces became same. 
Energy is decreased because system changes to get stability.
Potential at their surfaces are same,
So,
$V _1=V _2$
$\dfrac{Kq _1}{R _1}=\dfrac{Kq _2}{R _2}$
$q _1R _2=q _2R _1$
Hence Proved.

Two small spheres have mass ${m} _{1}$ and ${m} _{2}$ and hanging from massless insulating threads of lengths ${l} _{1}$ and ${l} _{2}$. Two spheres carry charges ${q} _{1}$ and ${q} _{2}$ respectively. The spheres hang such that they are on the same horizontal level and the threads are inclined to the vertical at angle ${\theta} _{1}$ and ${\theta} _{2}$ respectively. If $F _1 = F _2$, then:

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

  1. ${ \theta } _{ 1 }={ \theta } _{ 2 }$

  2. ${ M } _{ 1 }={ M } _{ 2 }$

  3. $\cfrac { l _{ 1 } }{ \tan { { \theta } _{ 1 } } } =\cfrac { l _{ 2 } }{ \tan { { \theta } _{ 2 } } } \quad $

  4. $\cfrac { q _{ 1 } }{ \tan { { \theta } _{ 1 } } } =\cfrac { q _{ 2 } }{ \tan { { \theta } _{ 2 } } } $

Show answer
Correct Option: B
Explanation:

For sphere $1$

In equilibrium, from figure.
$\begin{array}{l} { T _{ 1 } }\cos { \theta _{ 1 } } ={ M _{ 1 } }g; \ { T _{ 1 } }\sin { \theta _{ 1 } } ={ F _{ 1 } } \ \therefore \tan { \theta _{ 1 } } =\dfrac { { { F _{ 1 } } } }{ { { M _{ 1 } }g } } . \end{array}$
For sphere $2$
In equilibrium, from figure.
$\begin{array}{l} { T _{ 2 } }\cos { \theta _{ 2 } } ={ M _{ 2 } }g; \ { T _{ 2 } }\sin { \theta _{ 2 } } ={ F _{ 2 } } \ \therefore \tan { \theta _{ 2 } } =\dfrac { { { F _{ 2 } } } }{ { { M _{ 2 } }g } } . \end{array}$
Force of repulsion between two charges are same
$\therefore F _1=F _2$
$\theta _1=\theta _2$ only if $\dfrac{{{F _1}}}{{{M _1}g}} = \dfrac{{{F _2}}}{{{M _2}g}}.$
But $F _1=F _2$, then $M _1=M _2$.

Two particles $X$ and $Y$ having equal charges after being accelerated thorough the same potential difference enter a region of uniform magnetic field and describe circular paths of radius $R _1$ and $R _2$ respectively. the ratio of mass of $X$ to that of $Y$ is

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

  1. $\sqrt { R _1{/R _2} }$

  2. $R _2{/R _1}$

  3. $(R _1{/R _2})^2$

  4. $R _1{/R _2}$

Show answer
Correct Option: C
Explanation:

Given that,

The radius of particle X$={{R} _{1}}$

The radius of particle Y$={{R} _{2}}$


We know that,

Work done of each particle = $qV$


Now, suppose the particle starts from rest, and final kinetic energy is

For, X particle,

$\dfrac{1}{2}{{m} _{1}}v _{1}^{2}=qV$

For, Y particle

$\dfrac{1}{2}{{m} _{2}}v _{2}^{2}=qV$


Now,

${{m} _{1}}v _{1}^{2}={{m} _{2}}v _{2}^{2}.....(I)$

Now, from the magnetic force is

$ {{F} _{{{m} _{1}}}}=q{{v} _{1}}B $

$ {{F} _{{{m} _{2}}}}=q{{v} _{2}}B $


Now, the magnetic force is equal to the centripetal force is

For, X

$ \dfrac{{{m} _{1}}v _{1}^{2}}{{{R} _{1}}}=qB{{v} _{1}} $

$ {{m} _{1}}{{v} _{1}}=qB{{R} _{1}} $

For, Y

$ \dfrac{{{m} _{2}}v _{2}^{2}}{{{R} _{2}}}=qB{{v} _{2}} $

$ {{m} _{2}}{{v} _{2}}=qB{{R} _{2}} $


Now, putting the value in equation (I)

$ {{m} _{1}}{{\left( \dfrac{qB{{R} _{1}}}{{{m} _{1}}} \right)}^{2}}={{m} _{2}}{{\left( \dfrac{qB{{R} _{2}}}{{{m} _{2}}} \right)}^{2}} $

$ \dfrac{R _{1}^{2}}{{{m} _{1}}}=\dfrac{R _{2}^{2}}{{{m} _{2}}} $

$ \dfrac{{{m} _{1}}}{{{m} _{2}}}={{\left( \dfrac{{{R} _{1}}}{{{R} _{2}}} \right)}^{2}} $

 Hence, the ratio of the mass is ${{\left( \dfrac{{{R} _{1}}}{{{R} _{2}}} \right)}^{2}}$

 

An electron in a picture tube of TV set is accelerated from rest through a potential difference of $5\times 10^3V$.Then the speed of electron as a result of acceleration is going to be

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

  1. $1.2 \times 10^7m/s$

  2. $2.2\times 10^7m/s$

  3. $3.2 \times 10^7m/s$

  4. $4.2\times 10^7m/s$.

Show answer
Correct Option: D
Explanation:

By energy conservation 
$\dfrac{1}{2}mv^2=qv$
$\dfrac{1}{2}mv^2=1.6\times 10^{-19}\times 5\times 10^3$
$v^2=\dfrac{2\times 1.6\times 10^{19}\times 5\times 10^3}{9.1\times 10^{-31}}$
$v^2=0.175\times 10^{16}$
$v=0.419\times 10^8$
$=4.19\times 10^{7}$
$=4.2\times 10^7m/s$.

A charge $10$ esu is placed at a distance of $2$ cm, from a charge$ 40$ esu and $4$ cm. from another charge -$20$ esu. The potential energy of the charge $10$ esu is :- (In ergs)

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

  1. $87.5$

  2. $112.5$

  3. $150$

  4. $zero$

Show answer
Correct Option: A

Calculate the electrostatic potential energy of two electrons separated bt 3 $\overset { \circ  }{ A } $ in vacuum

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

  1. $7.69\times10^{-19} J$

  2. $8.69\times10^{-19} J$

  3. $5.69\times10^{-19} J$

  4. $6.69\times10^{-19} J$

Show answer
Correct Option: B

A proton and a deuteron are initialy at rest an dare accelerated through the same potential difference. Which following is false concerning  the final properties of the two particles?

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

  1. They have different speeds

  2. They have same momentum

  3. They have same kinetic energy

  4. They have been subjected to same force

Show answer
Correct Option: C

A flat circular fixed disc has a charge +Q uniformly distributed on the disc. A charge +q is thrown with kinetic energy K,towards the disc along its axis The charge is q

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

  1. will not hit the disc at the center

  2. may return back along its path after touching the disc

  3. may return back along its path without touching the disc

  4. any of the above three situation is possible depending on the magnitude of K

Show answer
Correct Option: A

Particle A having positive charge is moving directly head on towards initially stationary positively charged particle B. At the instant when A and B are closest together. 

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

  1. the momenta of A and B must be equal

  2. the velocities of A and B must be equal

  3. B would have gained less kinetic energy than A would have lost

  4. B would have gained the same momentum as A would have lost

Show answer
Correct Option: C,D

Eight charges (each $q$) are placed at the vertices of a regular cube of side $a$. The electric potential energy of the configuration will be $ U=12\times \dfrac { 1 }{ 4\pi \varepsilon _{ 0 } } ,\dfrac { q^{ 2 } }{ a } \times \quad x $ then x.

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

  1. $ 1+\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 3 } } $

  2. $ 1+\dfrac { 2 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 3 } } $

  3. $ 1+\dfrac { 2 }{ \sqrt { 2 } } +\dfrac { 2 }{ \sqrt { 3 } } $

  4. $ \left[ 1+\dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ 3\sqrt { 3 } } \right] $

Show answer
Correct Option: D

Two point charges of +10 $\mu c$ and -10 $\mu c$ are placed at a distance $40$ cm in air. Potential energy of the system will be-

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

  1. $2.25 J$

  2. $2.35 J$

  3. $-2.25 J$

  4. $-2.35 J$

Show answer
Correct Option: C

A solid non-conducting sphere of radius $R$ having charge density $\rho = \rho _{0}x$, where $x$ is distance from the centre of sphere. The self potential energy of the sphere is

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

  1. $\dfrac {\pi \rho _{0}^{2} R^{4}}{6\epsilon _{0}}$

  2. $\dfrac {\pi \rho _{0}^{2} R^{6}}{4\epsilon _{0}}$

  3. $\dfrac {\pi \rho _{0}^{2} R^{6}}{6\epsilon _{0}}$

  4. None of these

Show answer
Correct Option: C

If a proton and an electron are accelerated through the same potential difference:

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

  1. both the proton and electron have same K.E

  2. both the proton and electron have same momentum

  3. both the proton and electron have same velocity

  4. both the proton and electron have same temperature

Show answer
Correct Option: A
Explanation:

P.D = V
K.E = eV
Velocity will be equal and opposite

What is the change in potential energy of a particle of charge +q that is brought from a distance of 3r to a distance of 2r by a particle of charge q? 

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

  1. $kq^2/r$

  2. $-kq^2/6r$

  3. $kq^2/r^2$

  4. $-kq^2//4r^2$

  5. $8kq^2/r^2$

Show answer
Correct Option: B
Explanation:

As the $+q$ is brought near by charge $q$ so there is an attraction force it means q is a negative charge, equal to $-q$.
 The electric potential energy of a system of two point charges (  $+q$ ,  $-q$) sepatrated by a distance $3r$ is given by ,
                  $U=k\dfrac{q\times -q}{3r}$  , 
when the separation is reduced to $2r$ then potential energy will become ,
                 $U'=k\dfrac{q\times- q}{2r}$ 
therefore change in potential energy is 
                             $\Delta U=U'-U$ 
or                         $\Delta U=-k\dfrac{q\times q}{2r}+k\dfrac{q\times q}{3r}$
or                         $\Delta U=kq^{2}\left(1/3r-1/2r\right)=-kq^{2}/6r$

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by  potential difference of 100 V is:

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

  1. $\sqrt{\dfrac{m _{\alpha}}{m _e}}$

  2. $\sqrt{\dfrac{m _e}{m _{\alpha}}}$

  3. $\dfrac{2m _e}{m _{\alpha}}$

  4. $\sqrt{\dfrac{m _e}{2m _{\alpha}}}$

Show answer
Correct Option: D
Explanation:

Mass of proton $=1$ $u$

and, Mass of alpha particle $=4$ $u$
Charge of proton $=e$
And, charge of alpha particle $=2e$
For charge moving in a uniform potential field.
$K.E=$ Electric potential energy
$\cfrac{1}{2}mv^2=qV$
$v=\sqrt{\cfrac{2qV}{m}}$
OR,
Momentum $(P)=mv$
$\Rightarrow P=m\sqrt{\cfrac{2qV}{m}}$
$\Rightarrow P=\sqrt{m2qV}$
Momentum of proton,
$P _1=\sqrt{2qVm _1}=\sqrt{2eVm _e}$
Momentum of alpha particle 
$P _2=\sqrt{2qVm _2}=\sqrt{2\times 2eVm _a}$
$\therefore \cfrac{P _1}{P _2}=\sqrt{\cfrac{2eVm _e}{4eVm _a}}$
$\therefore \cfrac{P _1}{P _2}=\sqrt{\cfrac{m _e}{2m _a}}$

A sphere of radius $1$ cm has potential of $8000$V. The energy density near the surface of sphere will be?

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

  1. $64\times 10^5$ $J/m^3$

  2. $8\times 10^3$ $J/m^3$

  3. $32$ $J/m^3$

  4. $2.83$ $J/m^3$

Show answer
Correct Option: D
Explanation:
Energy density = $=(\frac { 1 }{ 2 } )∈0E2$ 
$=(\frac { 1 }{ 2 } )\times 8.86\times 10-12\times \left( \frac { v }{ r }  \right) 2$
$=4.43\times 10-12\times [(8000)/(10-2)]2$
$=283.5\times 10-2$
$=2.83J/m3$

Two unlike charges of magnitude q are separated by a distance 2d. The potential at a point midway between them is

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

  1. zero

  2. $\dfrac{1}{4 \pi {\epsilon} _{0}}$

  3. $\dfrac{1}{4 \pi {\epsilon} _{0}}$ . $\dfrac{q}{d}$

  4. $\dfrac{1}{4 \pi {\epsilon} _{0}}$ . $\dfrac{2q}{d}$

Show answer
Correct Option: A

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. $-2 \varepsilon _0$

Show answer
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

Show answer
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. $0.2\ m$

Show answer
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.
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