Tag: electrostatics

Questions Related to electrostatics

$100J$ 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. $zero$

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

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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

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Correct Option: C