Q: Two charges of +200\mu C and -200\mu C are placed at the corners B and C of an equilateral triangle ABC of side 0.1 m . The force on a charge of 5\mu C placed at A is :

\rightarrow The vertical components of force cancel out as the magnitude of charges are equal.Net force experience by A is F_{net}=\dfrac {k(200\times 10^{-6})(5\times 10^{-6})}{(0\cdot 1)^2}\times Cos (60^0)+\dfrac {k(200\times 10^{-6})(5\times 10^{-6})}{(0\cdot 1)^2}(Cos 60) =\dfrac {(9\times 10^9)(200\times 10^{-6})(5\times 10^{-6})}{10^{-2}} =9\times 20\times 5 F_{net}=900 N (\hat x)

Q: Two charged balls of the same radius and weight suspended on threads of equal length are immersed into a liquid having density of d_{1} and a dielectric constant K . The density d of the material of the balls for the angles of divergence of the threads in the air and in the dielectric to be the same is:

When in air , for each ball , T cos\theta = Density \times volume \times g = dVg ( Vertical component of tension is balanced by weight ) T sin\theta = \dfrac{q^{2} }{4\pi \epsilon_0 ( 2l sin \theta) ^{2} } ( The horizontal component of tension is balanced by the Electrostatic Force ) tan\ \theta=\dfrac{\dfrac{q^{2}}{4\pi \epsilon_0 (2l\ sin\theta)^2}}{dVg} ----------- (1)Similarly ,when immersed in liquid , tan \theta = \dfrac{ \dfrac{ q^{2} }{4\pi K \epsilon_0 (2l sin \theta) ^{2} }}{Vg ( d-d_{1} )} ----------(2) ( Buoyant force, gravity and electrostatic force ) From (1) and (2) K(d-d_{1}) = d Hence , d = \dfrac{K d_{1} }{K-1}

Q: Two particles A and B, having opposite charges 2.0\times 10^{-6} C and -2.0\times 10^{-6} C, are placed at a separation of 1.0 cm. Calculate the electric field at a point on the axis of the dipole 1.0 m away from the centre.

The value of the dipole moment is : p=q\times d p=2\times 10^{-6}\times 0.01 =2\times 10^{-8} We know, Electric field at an axial point of the dipole =\dfrac{2Kp}{r^3} =\dfrac{2\times 9\times 10^9\times 2\times 10^{-8}}{(1\times 10^{-2})^3} =36\times 10^7N/C .

Q: A particle of mass m and charge -q enters the region between the two charged plates initially moving along x-axis with speed {v}_{x} as shown in figure. The length of the plate is L and a uniform electric field E is maintained between the plates. The vertical deflection of the particle at the far edge of the plate is

\textbf{Hint - } In the vertical direction initial velocity, u=0 , and horizontal velocity is constant. \textbf{Step 1: Finding acceleration} Acceleration (a) = \dfrac {F}{m} = \dfrac {qE}{m} .....(i) ( \because Electric force = F = q \times E, where q is the charge and E is the electron voltage) \textbf{Step 2: Calculate vertical deflection} Time taken to cross the field of charge ( t ) = \dfrac{distance}{velocity} = \dfrac{L}{{v}_{x}} ....(ii) Now using third equation of motion, s=ut+\dfrac{1}{2}a{t}^{2} \therefore Deflection in vertical is y=0+\dfrac{1}{2}\times\dfrac{qE}{m}\times{(\dfrac{L}{{v}_{x}})}^{2} (using (I) and (ii)) \therefore y = \dfrac{qE{L}^{2}}{2m{v}^{2}_{x}} \textbf{Hence, the correct option is (A)} \dfrac{qE{L}^{2}}{2m{v}^{2}_{x}}

Question 1

Two point electric charges of values  q  and  2q  are kept at a distance  d  apart from each other in air. A third charge  Q  is to be kept along the same line in such a way that the net force acting on  q  and  2q  is zero. Find the location of the third charge from charge  q .

Accepted Answer

Q  should be negative and it should be placed somewhere between  q  and  2q For net force on  q  to be zero:  \cfrac { kqQ }{ { x }^{ 2 } } =\cfrac { kq2q }{ { d }^{ 2 } } ...(i)\quad  For net force on  2q  to be zero:  \cfrac { k2qQ }{ { (d-x) }^{ 2 } } =\cfrac { kq2q }{ { d }^{ 2 } } ....(ii) From Equations (i) and (ii) we get \cfrac { 1 }{ { x }^{ 2 } } =\cfrac { 2 }{ { (d-x) }^{ 2 } } \Rightarrow x=\cfrac { d }{ 1+\sqrt { 2 }  }

Question 2

Two charges of  +200\mu C  and  -200\mu C  are placed at the corners B and C of an equilateral triangle ABC of side  0.1 m . The force on a charge of  5\mu C   placed at A is :

Accepted Answer

ightarrow   The vertical components of force cancel out as the magnitude of charges are equal.Net force experience by A is  F_{net}=\dfrac {k(200	imes 10^{-6})(5	imes 10^{-6})}{(0\cdot 1)^2}	imes Cos (60^0)+\dfrac {k(200	imes 10^{-6})(5	imes 10^{-6})}{(0\cdot 1)^2}(Cos 60) =\dfrac {(9	imes 10^9)(200	imes 10^{-6})(5	imes 10^{-6})}{10^{-2}} =9	imes 20	imes 5 F_{net}=900 N (\hat x)

Question 3

Two charged balls of the same radius and weight suspended on threads of equal length are immersed into a liquid having density of  d_{1}  and a dielectric constant  K . The density  d  of the material of the balls for the angles of divergence of the threads in the air and in the dielectric to be the same is:

Accepted Answer

When in air , for each ball ,  T cos	heta = Density 	imes volume 	imes g = dVg  ( Vertical component of tension is balanced by weight )  T sin	heta = \dfrac{q^{2} }{4\pi \epsilon_0 ( 2l sin 	heta) ^{2} }   ( The horizontal component of tension is balanced by the Electrostatic Force ) tan\ 	heta=\dfrac{\dfrac{q^{2}}{4\pi \epsilon_0 (2l\ sin	heta)^2}}{dVg}  ----------- (1)Similarly ,when immersed in liquid , tan  	heta = \dfrac{ \dfrac{  q^{2} }{4\pi K \epsilon_0  (2l sin 	heta) ^{2} }}{Vg ( d-d_{1} )}     ----------(2) ( Buoyant force, gravity  and electrostatic force ) From (1) and (2) K(d-d_{1}) = d Hence ,   d =    \dfrac{K d_{1} }{K-1}

Question 4

Two particles A and B, having opposite charges  2.0	imes 10^{-6} C and  -2.0	imes 10^{-6} C, are placed at a separation of  1.0 cm. Calculate the electric field at a point on the axis of the dipole  1.0 m away from the centre.

Accepted Answer

The value of the dipole moment is : p=q	imes d p=2	imes 10^{-6}	imes 0.01 =2	imes 10^{-8} We know, Electric field at an axial point of the dipole =\dfrac{2Kp}{r^3} =\dfrac{2	imes 9	imes 10^9	imes 2	imes 10^{-8}}{(1	imes 10^{-2})^3} =36	imes 10^7N/C .

Question 5

An electric dipole consists of two opposite charges of magnitude  1\mu C  separated by a distance of 2cm. The dipole is placed in an electric field  10^{-5}Vm^{-1} . The maximum torque that the field exerts on the dipole is:

Accepted Answer

The torque experienced by a dipole in an electric field E is   \overrightarrow p 	imes \overrightarrow E  Maximum torque  =pE=qaE = 10^{-6}	imes 0.02	imes 10^{-5} = 2	imes 10^{-13}\ Nm

Question 6

A dipole is placed in an electric field whose direction is fixed but its magnitude varies with distance. It is possible that the dipole experiences :

Accepted Answer

Dipole means two charges  +q  and  -q  are placed at a certain distance.There  three cases may be occured for electric field whose direction is fixed but magnitude varies with distance.1) when forces on the two charges are in different direction and different magnitude, they will have a net force and a torque.2) when forces on the two charges are in same direction but different magnitude, they will have a net force but no torque.3) when forces on the two charges are in opposite direction but same magnitude, they will have no net force but a torque.

Question 7

A particle of mass 1kg and carrying positive charge 0.01 C is sliding down an inclined plane of angle  30^{0} with the horizontal. An electric field E is applied to stop the particle. If the coefficient of friction between the particle an the surface of the plane is \dfrac{1}{2\sqrt{3}} , E must be:

Accepted Answer

In equilibrium N=E q sin 30+ mg cos 30-------(1)  and f+E q cos 30= mg sin 30-------(2) Maximum static function is  \mu N \Rightarrow f=\mu N \mu N+ Eq cos 30= mg sin 30-------(3) Solving (1), (2) and (3) -\dfrac {Eq cos 30}{\mu}+\dfrac {mg sin 30}{\mu}=Eq sin 30+mg cos 30 E(\dfrac {q cos 30}{\mu}+q sin 30)= mg(-cos 30+\dfrac {sin 30}{\mu}) E	imes (0\cdot 01)(\dfrac {\sqrt 3	imes 2\sqrt 3}{2}+\dfrac {1}{2})=1	imes 10(\dfrac {-\sqrt 3}{2}+\dfrac {1}{2}2\sqrt 3) E	imes 0\cdot 01	imes 5\dfrac {7}{2}=10	imes (\dfrac {\sqrt 3}{2}) E=\dfrac {10\sqrt 3}{0\cdot 01	imes 7}=\dfrac {1000\sqrt 3}{7}

Question 8

A particle of mass  m  and charge  -q  enters the region between the two charged plates initially moving along x-axis with speed  {v}_{x}  as shown in figure. The length of the plate is  L  and a uniform electric field  E  is maintained between the plates. The vertical deflection of the particle at the far edge of the plate is

Accepted Answer

extbf{Hint - }  In the vertical direction initial velocity,  u=0 , and horizontal velocity is constant. 	extbf{Step 1: Finding acceleration}  &#10;&#10;Acceleration (a) =  \dfrac {F}{m} &#8203;=  \dfrac {qE}{m}  .....(i)  ( \because   Electric force = F = q  	imes  E, where  q  is the&#10;charge and  E  is the electron voltage) 	extbf{Step 2: Calculate vertical deflection}   Time taken to cross the field of charge ( t ) =  \dfrac{distance}{velocity} = \dfrac{L}{{v}_{x}} &#8203;....(ii) Now using third equation of motion,   s=ut+\dfrac{1}{2}a{t}^{2} &#10;&#10; 	herefore  Deflection in vertical is&#10;&#10; y=0+\dfrac{1}{2}	imes\dfrac{qE}{m}	imes{(\dfrac{L}{{v}_{x}})}^{2}   (using (I) and (ii))&#10;&#10; 	herefore y  =&#8203;  \dfrac{qE{L}^{2}}{2m{v}^{2}_{x}} &#10;&#10; 	extbf{Hence,&#10;the correct option is (A)}  &#8203;  \dfrac{qE{L}^{2}}{2m{v}^{2}_{x}} &#10;&#10;

Question 9

A thin fixed ring of radius  1 \ m  has a positive charge  1	imes 10^{-5}C  uniformly distributed over it. A particle of mass  0.9gm  having a negative charge of  1	imes 10^{-6} C  is placed on the axis at a distance of  1 cm  from the centre of the ring. Assuming that the oscillations has small amplitude, the time period of oscillations is:

Accepted Answer

We know  E  on the axis of the ring is: E=\dfrac {k q x}{(r^2+x^2)^{3/2}} For small amplitudes  r>>x : \Rightarrow E=\dfrac {k q x}{r^3} We can see that force is always applied in the opposite direction of displacement i.e: F=\dfrac {-k q x}{r^3}(1	imes 10^{-6})C \Rightarrow a=\dfrac {-k q x (1	imes 10^{-6})}{m r^3} We know for a particle in S. H. M:  a=-\omega^2 x 	herefore \omega^2=\dfrac {k q (1	imes 10^{-6})}{m r^3}=\dfrac {9	imes 10^9	imes 1	imes 10^{-5}	imes 1	imes 10^{-6}}{0\cdot 9	imes 10^{-3}	imes 1^3}=100 \Rightarrow \omega=10 Now,  T=\dfrac {2\pi}{w}=\dfrac {2	imes 3\cdot 14}{10}=0.63s

Question 10

Charge of uniform volume density  \rho = 3.2 \mu_C/m^3   fills a nonconducting solid sphere of radius  5.0 cm . What is the magnitude of the electric field (a)  3.5 cm  and (b)  8.0 cm  from the sphere&#8217;s center?

Accepted Answer

Since the charge distribution is uniform, we can find the total charge q by multiplying  &#961;  by the spherical volume (  \frac{4}{3}&#960;r^3   ) with  r = R = 0.050  m. This gives  q = 1.68 nC . (a) Applying Eq. with   r = 0.035 m , we have   E_{internal}=\dfrac{|q|r}{4\pi \epsilon_0R^3} . (b) Outside the sphere we have (with r = 0.080 m)                                          E_{external}=\dfrac{|q|}{4\pi \epsilon_0r^2}=\dfrac{(8.99 	imes10^9 N.m^2/C^2 )(1.68	imes 10^{-9} C)}{(0.080 m)^2}=2.4	imes10^3\,N/C .

Question 11

Charge Q is distributed uniformly on length  l  of a wire. It is bent in the form of a semicircular ring. Find the electric field at the centre of the ring.

Accepted Answer

Consider two small elements of length  dl  each charge  dq  on each element. If  \lambda  is the line charge density,  Q=\lambda l=\lambda . \pi r . as wire is bent in the form of semicircular ring.Hence ,    dq=\lambda dl=\frac{Q}{\pi r}.(r d	heta)=\frac{Q}{\pi} d	heta  .  Here sine component of the field at center cancel out and  only cosine will contribute the filed.thus,  \displaystyle dE_c=2dE\cos	heta=2\frac{dq}{4\pi\epsilon_0 r^2}\cos	heta=\frac{Q}{2\pi^2 \epsilon_0 r^2}\cos	heta d	heta  or  \displaystyle E_c=\frac{Q}{2\epsilon_0(\pi r)^2}\int_0^{\pi/2}\cos	heta d	heta=\frac{Q}{2\epsilon_0(\pi r)^2}(\sin\frac{\pi}{2}-\sin0)=\frac{Q}{2\epsilon_0l^2}      as  (l=\pi r)

Question 12

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density  \lambda  without using Gauss's law. [Hint: Use Coulombs law directly and evaluate the necessary integral.]

Accepted Answer

Take a long thin wire of uniform linear charge density  \lambda .Consider a point A at a perpendicular distance  l  from the mid-point  O  of the wire, as shown in the following figure.Let  E  be the electric field at point  A  due to the wire,  XY .Consider a small length element  dx  on the wire section with  OZ = x Let  q  be the charge on this piece. 	herefore q=\lambda dx Electric field due to the piece, dE\displaystyle =\frac{1}{4\pi\in_0}\frac{\lambda dx}{(AZ)^2} However,  AZ=\sqrt{(l^2+x^2)} 	herefore dE \displaystyle =\frac{\lambda dx}{4\pi\in_0(l^2+x^2)} The electric field is resolved into two rectangular components.  dE\, cos\,	heta  is the perpendicular component and  dE\, sin\,	heta  is the parallel component.When the whole wire is considered, the component  dE\,sin\,	heta  is cancelled.Only the perpendicular component  dE\,cos\,	heta  affects point A.Hence, effective electric field at point A due to the element  dx  is  dE_1 . 	herefore dE_1=\displaystyle \frac{\lambda dx\, cos\,	heta}{4\pi\in_0(x^2+l^2)}   ... (1)In  	riangle AZO , tan\, 	heta = \frac{x}{l} x=l\, tan\,	heta   ... (2)On differentiating equation (2), we obtain dx=l\sec^2(	heta)d	heta dE_1=\cfrac{\lambda cos (	heta) d 	heta}{4 \pi \varepsilon_o l} \displaystyle E_1=\int_{-\pi/2}^{\pi/2} \frac{\lambda \cos(	heta) d 	heta}{4 \pi \varepsilon_o l}  Solving,  E_1=\dfrac{\lambda}{2 \pi \varepsilon_o l}

Question 13

A point charge  +10\ \mu C  is a distance  5\ cm  directly above the center of a square of side  10\ cm,  as shown in fig. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge  10\ cm. )

Accepted Answer

extbf{Step 1: Finding closed surface for application of Gauss Law}  The square can be considered as one face of a cube of edge  10 cm  with a centre where charge  q  is placed. According to Gauss&#8217;s law for a cube, total electric flux is through all its six faces. \phi_{total}=\dfrac{q}{\in_0} 	extbf{Step 2: Flux through one face or square} As the charge is placed symmetrically to each face of the cube, thus electric flux passing through each face is equal.So, Electric flux through one face of the cube i.e., through the square, \displaystyle \phi_1=\dfrac{\phi_{total}}{6} =\dfrac{1}{6}\dfrac{q}{\in_0} \Rightarrow \ \ \displaystyle \phi_1=\frac{10	imes 10^{-6}C}{6 	imes 8.854	imes 10^{-12}N^{-1} m^{-2} C^2} =1.88	imes 10^5Nm^2C^{-1} Therefore, electric flux through the square is  1.88	imes 10^5Nm^2C^{-1} .

Question 14

Two point charges  q  and  -q  are separated by a distance  2a  (Fig.). Evaluate the flux of electric field strength vector across a circle of radius  R .

Accepted Answer

Electric field due to a dipole at equatorial line E = \dfrac{2ql}{4\pi \varepsilon_0(l^2+x^2)^{3/2}} Let elemental surface area be  xd	heta  dx \displaystyle \phi = \int{Ex d	heta dx} = \dfrac{2ql}{4\pi\varepsilon_0}\int_{0}^{R} \dfrac{xdx}{(l^2+l^2)^{3/2}}\int_{0}^{2\pi}d	heta =\displaystyle \dfrac{ql}{\varepsilon_0}\int_{0}^{R}\dfrac{xdx}{(l^2+x^2)^{3/2}}=\dfrac{q}{\varepsilon_0}\left[1-\dfrac{l}{\sqrt{l^2+R^2}}ight]

Question 15

It has been experimentally observed that the electric field in a large region of Earth's atmosphere is directed vertically down. At an altitude of  300\ m , the electric field is  60\ Vm^{-1} . At an altitude of  200\ m , the field is  100 Vm^{-1} . Calculate the net amount of charge contained in the cube of  100\ m  edge, located between  200  and  300\ m  altitude.

Accepted Answer

Given: Cube of edge, 100m located between 200 and 300 m altitude. At altitude 300m, E=  60Vm^{-1} , and at 200 m, E=  100Vm^{-1} To find: the net amount of charge contained in the cube Sol:We will use Gauss' law. The net flux through the cube is proportional to the charge contained within the cube.As the field is vertical, there is only flux through the horizontal faces. The upper face has a field of  60Vm^{-1} or\ 60NC^{-1}  across an area of  10,000 m^2  &#8211; so the total flux through this surface is  600 000 Nm^2C^{-1}  (which we&#8217;ll take as negative because it goes into the cubical volume). Similarly, the flux through the lower surface  1 000 000 Nm^2C^{-1} .The net flux through the surface then is  400 000 Nm^2C^{-1} By Gauss' Law, we know that the net flux through the surface is related to the internal charge by the formula: Flux_{net} = \dfrac {Q_{tot}}{ \epsilon_0} Q_{tot} = Flux_{net} 	imes \epsilon_0 = 400 000 	imes 8.8542 	imes 10^{-12} Q_{tot} = 3.54 	imes 10^{-6} C is the required net amount of charge contained in the cube.

Question 16

Consider the Gaussian surface that surrounds part of the charge distribution shown in Fig. Then, the contribution to the electric field at point  P  arises from charges.

Accepted Answer

Electric field at any point on Gaussian surface is due to all charges present inside or outside the surface.

Question 17

The figure shows the electric field lines of force emerging from a charged body. If the electric field at A and B are  E_{A}  and  E_{B}  respectively and if the distance between A and B is 'r' then :

Accepted Answer

In the case of an electric field diagram, the direction of the electric field can be found out by seeing the direction of the electric field lines. However, the magnitude of the electric field can be compared by comparing the closeness and clustering of electric field lines at the point.At point A, the lines are more closely spaced than at B. Therefore,  E_A  >  E_B

Question 18

An uncharged metal sphere is placed between two equal and oppositely charged metal plates. The nature of lines of force will be

Accepted Answer

\bullet  Electric field lines should be perpendicular to equipotential surface so here field lines should be perpendicular to metal surface because potential of all the points of metal is equal.Hence option B correct

Electric Charges And Fields