Tag: potential energy of a system of charges

Quizzes (0) Questions (29)

Questions Related to potential energy of a system of charges

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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:

  1. 50 V

  2. 500 V

  3. 5 V

  4. 5000 V
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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_______

  1. zero

  2. positive

  3. negative

  4. zero if the charge Q is at the center and nonzero otherwise.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The net displacement round one complete circle is 0.

So, the work done is 0.

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

$100J$ 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

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

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

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

  4. $5\times10^{7}V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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:

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $10\;volts.$
Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

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 $R _2$.

  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. $\dfrac{q^2}{8\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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).
Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

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:

  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 ${q} _{1}{R} _{2}={q} _{2}{R} _{1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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:

  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 } } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $M _1=M _2$.

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

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

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

  2. $R _2{/R _1}$

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

  4. $R _1{/R _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 ${{\left( \dfrac{{{R} _{1}}}{{{R} _{2}}} \right)}^{2}}$

 

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

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

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

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

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

  4. $4.2\times 10^7m/s$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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