Physics

Electrostatics

299 Questions

Electrostatics deals with electric charges, fields, and potentials at rest. It is a crucial topic for physics sections in engineering and civil services competitive examinations. Review these questions to build a strong understanding of Coulomb law, electric dipoles, Gauss law, and electric flux.

Electric field and potentialElectric dipole momentGauss Law applicationsCoulomb force calculationsCharge distribution on spheresEquipotential surfaces

Electrostatics Questions

Multiple choice capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

Inside a hollow charged spherical conductor, the electric field is found to be.

  1. Proportional to the distance from the centre

  2. A function of the area of the sphere

  3. Zero

  4. A function of the charge density of the sphere
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Gauss's Law, the electric field inside a hollow charged conductor is zero because there is no enclosed charge.

Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

Two identical conducting balls having positive charges $q _1$ and $q _2$ are separated by a distance r. If they are made to touch each other and then separated to the same distance, the force between them will be

  1. less than before

  2. same as before

  3. more than before

  4. zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If they are made to touch each other, the electrostatic conduction will occur. As a result the total charge will get distributed equally on both the balls because both the balls are identical in nature. 


Initially, $F \propto q _1q _2$      (Since $F = \dfrac{kq _1q _2}{r^2}$)


Now after making contact, charge on each ball will become $Q _1=Q _2=\dfrac{q _1+q _2}{2}$
Now, $F' \propto (\dfrac{q _1 + q _2}{2})^2$
(Because they are separated to the same distance) 


From above we can say that $F' > F$

Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

 A conducting wire is connected between two conducting spheres of equal size have a charge of -3C and +1C respectively. Find out the new charge on each sphere ? 

  1. -4C

  2. +4C

  3. -1C

  4. +1C

  5. Zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the two spheres are connected by a wire, the charge will flow between them till the potential between them will be same. Thus, the charge on each sphere is the average charge i.e $Q _{av}=\dfrac{-3+1}{2}=-1 C$

Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

The linear charge density of a thin metallic rod varies with the distance $'x'$ from one end as $\lambda  = {\lambda _0}{x^2}\left( {0 \leqslant x \leqslant l} \right).$ The total charge on the rod is:

  1. $\dfrac{{{\lambda _0}{l^3}}}{3}$
  2. $\dfrac{{{\lambda _0}{l^4}}}{3}$
  3. $\dfrac{{2{\lambda _0}{l^3}}}{3}$
  4. $\dfrac{{{\lambda _0}{l}}}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\lambda =\dfrac {\lambda _0 x}{L}$
$Q=\displaystyle \int _0^L \lambda \ dx$
$=\displaystyle \int _0^L \dfrac {\lambda _0 x}{L}dx$
$=\dfrac {\lambda _0}{L} \dfrac {x^2}{2}\displaystyle \int _0^L$
$=\dfrac {\lambda _0}{L}\times \dfrac {L^2}{2}$
$=\dfrac {\lambda _0L}{2}$
Multiple choice physics static electricity exchange of electrons methods used for charging electric charge

A hollow metal sphere, Sphere A, sits on an insulating stand. Sphere A has a diameter of 4 inches, and a net charge of magnitude $Q _0$. A second hollow metal sphere, Sphere B, also sits on an insulating stand, but has a diameter of 8 inches and zero net charge. The two spheres are brought close so that they touch, then they are separated.
In terms of $Q _0$, what is the final charge on Sphere A?

  1. $\cfrac{Q _0}{5}$
  2. $\cfrac{Q _0}{4}$
  3. $\cfrac{Q _0}{2}$
  4. $Q _0$
  5. $4Q _0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$R _B=2R _A$


Let final charge on spheres be $Q _A$ and $Q _B$.

On touching the spheres charge density becomes same.

And, charge density, $\sigma=\dfrac{Q}{4\pi R^2}$

$\implies Q\propto R^2$ for same $\sigma$

Hence, $\dfrac{Q _B}{Q _A}=\dfrac{R _B^2}{R _A^2}=2^2$

$\implies Q _B=4Q _A$

And, total charge$=Q _B+Q _A=Q _o$

$\implies 4Q _A+Q _A=Q _o$

$\implies Q _A=\dfrac{Q _o}{5}$

Answer-(A)

Multiple choice physics option a: relativity maxwell's equations the nature of light introduction to electromagnetic waves

The electric field associated with an e.m. wave in vacuum is given by $\vec {E} = 40\cos (kz - 6\times 10^{8}t)\hat {i}$, where $E, z$ and $t$ in $volt/m$, meter and seconds respectively. The value of wave vector $k$ is

  1. $6m^{-1}$
  2. $3m^{-1}$
  3. $2m^{-1}$
  4. $0.5m^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given: The electric field associated with  an electromagnetic wave in vacuum is given by $\vec E =40 \cos(kz−6\times 10^8t)\hat i$  , where E, z and t are in volt per meter, meter and second respectively.
To find the value of wave vector k
Solution: 
We know electromagnetic wave eqution is
$E=E _0\cos(kz-\omega t)$
And given equation is
$\vec E =40 \cos(kz−6\times 10^8t)\hat i$
By comparing these two, we get
$\omega=6\times10^8$ and 
$E _0=40\hat i$
we also know,
Speed of electromagnetic wave, $v=\dfrac \omega k$
where v is the speed of the light
Hence, $k=\dfrac \omega v\\\implies k=\dfrac {6\times 10^8}{3\times 10^8}\\\implies k=2m^{-1}$
is the required value
Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

A small charged ball of mass m and charge q is suspended from the highers point of a ring of radius R by means of an insulated code of negligible mass.The ring is made of a rigid wire of negligible cross-section and lies in a vertical plane.On the ring, there is uniformly distributed charge Q of the same as that of q .determine the length of the cord so as the equilibrium position of the ball lies on the symmetry axis ,perpendicular to the plane of the ring. 

  1. $\left( \cfrac { 2kQqR }{ mg } \right) ^{ 1/3 }$
  2. $\left( \cfrac { kQqR }{ mg } \right) ^{ 1/3 }$
  3. $\left( \cfrac { kQqR }{ 2mg } \right) ^{ 1/3 }$
  4. $\left( \cfrac { kQqR }{ mg } \right) ^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

A uniform electric field 'E' is directed towards positive X-axis. If at X=0, the electric potential is zero, then the potential at $X=+X _0,$ would be 

  1. $\dfrac{E}{X _0}$
  2. $\dfrac{-E}{X _0}$
  3. $-EX _0$
  4. $EX _0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that electric potential $V=\dfrac { dE }{ dx } $

Since $X=x+{ x } _{ 0 }$  so,
$V=\dfrac { E }{ { x } _{ 0 } } $

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

The path of cathode rays in an electric field can be approximated to a circle of radius r. In order to double the radius of the circular path, we must 

  1. reduce the electric field to half

  2. double the electric field

  3. increase electric field four times

  4. reduce electric field to one fourth
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, path of cathode rays is nearly circular. So, electric force must be acting as centripetal force for the cathode rays and it is given by
$ F _e = F _c $
qE = $ m \omega ^2 r$
we can conclude that $E \alpha r$ ,provided all the other quantities are constant.
Thus, to double the radius we should double electric field strength.

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

A positive charge is released from the origin at a place where uniform electric field $E$ and a uniform magnetic field be exist along the positive $y-$axis and positive $z-$axis respectively, then :  

  1. initial the charge particle tends to move along positive $z-$axis
  2. initial the charge particle tends to move along negative $Y-$direction
  3. initial the charge particle tends to move along positive $y-$direction
  4. the charge particle moves in $y-z$ plane
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Lorentz force is F = q(E + v x B). Initially, the velocity v is zero, so the magnetic force is zero. The electric field E is along the positive y-axis, so the initial force F = qE is also along the positive y-axis, causing the particle to accelerate in that direction.

Multiple choice motion of charged particle in magnetic field and electric field moving charges and magnetism magnetic effects of current and magnetism physics

Three equal charges, each having a magnitude of $ 4 \mu C$ , are placed at the three corners of a right-angled triangle of sides $6 cm, 8 cm$ and $10 cm.$ The force on the charge at the right-angle corner will be

  1. $ 11.5 N $
  2. $23 N $
  3. $46 N $
  4. $230 N $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two forces are,

${F _1} = \dfrac{{k{Q _1}{Q _2}}}{{{R _1}^2}}$

$ = \dfrac{{9 \times {{10}^9} \times {{\left( {4 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {6 \times {{10}^{ - 2}}} \right)}^2}}}$

$ = 10{\rm{N}}$

${F _2} = \dfrac{{k{Q _1}{Q _2}}}{{{R _2}^2}}$

$= \dfrac{{9 \times {{10}^9} \times {{\left( {4 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {8 \times {{10}^{ - 2}}} \right)}^2}}}$

$= \dfrac{{360}}{{64}} = 5.62{\rm{N}}$

Resultant force at the right angle vertex is,

$F = \sqrt {\left( {{F _1}^2 + {F _2}^2} \right)} $

$= \sqrt {{{10}^2} + {{5.62}^2}} $

$ = \sqrt {131.56} $

$= 11.5{\rm{N}}$

Multiple choice physics option a: relativity the nature of light speed of light and optical density introduction to light

The electric field associated with an electromagnetic wave in vacuum is given by $|\overrightarrow { E } |= 40\ cos (kz -6\times{10}^{8}t )$, where $E$, $z$ and $t$ are in volt per meter, meter and second respectively. The value of wave vector $k$ is:

  1. $2\ {m}^{-1}$
  2. $0.5\ {m}^{-1}$
  3. $3\ {m}^{-1}$
  4. $6\ {m}^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: The electric field associated with  an electromagnetic wave in vacuum is given by $|\vec E|=40 \cos(kz−6\times 10^8t)$


To find: Value of wave vector $k$


Solution: 
We know electromagnetic wave eqution is
$|\vec E|=E _0\cos(kz-\omega t)$

And given equation is
$|\vec E|=40 \cos(kz−6\times 10^8t)$

By comparing these two, we get
$\omega=6\times10^8$ and 
$E _0=40$

We also know,
Speed of electromagnetic wave is given by:
$v=\dfrac \omega k$
where v is the speed of the light.

Hence, 
$k=\dfrac \omega v\\\implies k=\dfrac {6\times 10^8}{3\times 10^8}\\\implies k=2m^{-1}$

Option $(A)$ is correct.