Q: Two charges 5\mu C and -20\mu C are placed 30 cm apart. Find the strength of electric field at the mid point between them.

Let "O" be the midpoint between the charges Q_1 and Q_2 .The strength of electric field at the point 'O' due to Q_1 is. E_1=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q_1}{d_{1}^{2}} =9\times 10^9\times \dfrac {5\times 10^{-6}}{(0.15)^2} =2\times 10^6N The strength of electric field at the point 'O' due to Q_1 is, E_2=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q_2}{d_{2}^{2}}=9\times 10^9\times \dfrac {20\times 10^{-6}}{(0.15)^2} =8\times 10^6N Total strength of electric field at the point 'O' s E=E_1+E_2 =2\times 10^6+8\times 10^6=[2+8]10^6=10^7N

Q: An electric dipole with dipole moment 4\times 10^{-9} C m is aligned at 30^\circ with the direction of a uniform electric field of magnitude 5\times 10^{4} NC^{-1} . Calculate the magnitude of the torque acting on the dipole.

Electric dipole moment, E = 4 \times 10^{-9} N/C Angle made by p with a uniform electric field, θ = 30°Electric field, E = 5 \times 10^4 N/C Torque acting on the dipole: \tau=pE\sin \theta =4\times 10^{-9}\times 5\times 10^4\times sin\, 30^o =20\times 10^{-5}\times \frac{1}{2} =10^{-4}Nm Hence, the magnitude of the torque acting on the dipole is 10^{-4} N m.

Q: Three equal changes 2.0 \times 10 ^ {-6} C each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two.

Q = 2\times 10^{-6}C To find-Coulomb force experienced by one of the charge due to other two changes |\vec{F_1}| = \dfrac{Kq^2}{a^2} = F |\vec{F_2}| = \dfrac{Kq^2}{a^2} = F F_{Net} = \sqrt{F_!^2 + F_2^2+2F_1F_2\cos \theta} =\sqrt{F^2+F^2+2F^2\cos 60} = \sqrt{3}F =\sqrt{3}\times \dfrac{Kq^2}{a^2} = \dfrac{\sqrt{3}\times 9 \times 10^9\times (2\times 10^{-6})^2}{(5\times 10^{-2})^2} =\dfrac{\sqrt{3}\times 9\times 10^9\times 4\times 10^{-12}}{25\times 10^{-4}} = \dfrac{\sqrt{3} \times 9\times 4\times 10}{25} = 24.94N

Question 1

Define one "Coulomb" on the basis of Coulomb's law.

Accepted Answer

One coulomb is defined as the quantity of charge, which when placed at a distance of  1  metre in air or vacuum from an equal and similar charge, experiences a repulsive force of  9	imes { 10 }^{ 9 }N .

Question 2

A solid metallic sphere is placed in a uniform electric field. Which of the curves shown in figure represent the lines of force correctly?

Accepted Answer

Consider a Gaussian surface inside the metallic sphere. The charge enclosed by the Gaussian surface is zero (since metallic sphere is a conductor and charge lies only on the surface). Therefore, electric flux is zero.Now, Electric Flux = Electric Field * Areabut area cannot be zero. Therefore Electric field inside the sphere is zero. If electric field had any&#10;component parallel to the surface of the conductor, then this would cause surface charges&#10;to move. Therefore the&#10;electric field must lie perpendicular to the surface everywhere.

Question 3

Two large parallel planes charged uniformly with surface charge densities  \sigma  and  -\sigma  are located as shown in the figure. Which one of the following graphs shows the variation of electric field along a line perpendicular to the planes as one moves from A to B?

Accepted Answer

Electric field by a plane of charge density   \sigma   is given by  \dfrac {\sigma}{2\epsilon_0}  Thus,In the region I and III net electric field is zero.In the region II net electric field is \dfrac {\sigma}{2\epsilon_0}+\dfrac {\sigma}{2\epsilon_0}=\dfrac {\sigma}{\epsilon_0}  and is uniform.

Question 4

Two charges  5\mu C  and  -20\mu C  are placed 30 cm apart. Find the strength of electric field at the mid point between them.

Accepted Answer

Let &#34;O&#34; be the midpoint between the charges  Q_1 and Q_2 .The strength of electric field at the point 'O' due to  Q_1  is. E_1=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q_1}{d_{1}^{2}} =9	imes 10^9	imes \dfrac {5	imes 10^{-6}}{(0.15)^2} =2	imes 10^6N The strength of electric field at the point 'O' due to  Q_1  is, E_2=\dfrac {1}{4\pi\epsilon_0}\dfrac {Q_2}{d_{2}^{2}}=9	imes 10^9	imes \dfrac {20	imes 10^{-6}}{(0.15)^2} =8	imes 10^6N Total strength of electric field at the point 'O' s  E=E_1+E_2 =2	imes 10^6+8	imes 10^6=[2+8]10^6=10^7N

Question 5

In a Millikan's oil drop method, a charged drop of oil has mass m and charge q. It is made stationary by applying an electric field E. Find the strength of the electric field. If the result of Millikans experiment with different drops is  6.48 	imes 10^{-19} C, 12.82 	imes 10^{-19} C, 19.3 	imes 10^{-19}C, 8.02 	imes 10^{-19}C, 25.62 	imes 10^{-19}C, 16.04 	imes 10^{-19}C,  find the elementary charge.

Accepted Answer

If suspended charged oil drop has mass  m  and charge  q , then its weight  m g  is exactly equal to the electric force applied  q E , where  E  is the electric field.i.e.,   mg = q E \implies E = \dfrac {mg}{q} Also, irrespective of the charge on different oil drops, the elementary charge is always a multiple of the charge of an electron ( -1.6 	imes 10^{-19} ).

Question 6

If  E_{a}  be the electric field intensity due to a short dipole at a point on the axis and  E_{r}  be that on the right bisector at the same distance from the dipole, then

Accepted Answer

If  q  be the charge on one point of the dipole and the half length is equal to  d ,then the total dipole moment is given by,  p   =  2qd If   {E}_{a}   be the field at any axial point, then it is given by  \displaystyle \frac{kp}{{r}^{3}}  Similarly, the field at any point on the right bisector is given  \displaystyle {E}_{r} =  \frac{kp}{2{r}^{3}}  Hence,   {E}_{a} = 2 {E}_{r}

Question 7

An electric dipole with dipole moment  4	imes 10^{-9} C m is aligned at  30^\circ  with the direction of a uniform electric field of magnitude  5	imes 10^{4} NC^{-1} . Calculate the magnitude of the torque acting on the dipole.

Accepted Answer

Electric dipole moment,   E = 4 	imes 10^{-9} N/C Angle made by p with a uniform electric field, &#952; = 30&#176;Electric field,   E = 5 	imes 10^4 N/C Torque acting on the dipole:  	au=pE\sin 	heta =4	imes 10^{-9}	imes 5	imes 10^4	imes sin\, 30^o =20	imes 10^{-5}	imes \frac{1}{2} =10^{-4}Nm Hence, the magnitude of the torque acting on the dipole is  10^{-4}  N m.

Question 8

Three equal changes  2.0 	imes 10 ^ {-6} C  each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two.

Accepted Answer

Q = 2	imes 10^{-6}C To find-Coulomb force experienced by one of the charge due to other two changes |\vec{F_1}| = \dfrac{Kq^2}{a^2} = F |\vec{F_2}| = \dfrac{Kq^2}{a^2} = F F_{Net} = \sqrt{F_!^2 + F_2^2+2F_1F_2\cos 	heta} =\sqrt{F^2+F^2+2F^2\cos 60} = \sqrt{3}F =\sqrt{3}	imes \dfrac{Kq^2}{a^2} = \dfrac{\sqrt{3}	imes 9 	imes 10^9	imes (2	imes 10^{-6})^2}{(5	imes 10^{-2})^2} =\dfrac{\sqrt{3}	imes 9	imes 10^9	imes 4	imes 10^{-12}}{25	imes 10^{-4}} = \dfrac{\sqrt{3} 	imes 9	imes 4	imes 10}{25} = 24.94N

Question 9

The electric field in a region is radially outward with magnitude  E=\alpha r . Calculate the charge contained in a sphere of radius R centered at the origin. Calculate the value of the charge if  \alpha =100 Vm^{-2}  and R=0.30 m.

Accepted Answer

Consider a spherical shell of radius x. The electric flux through this surface is  \phi= \int_{s}\vec{E}\cdot d\vec{S}=E_r4 \pi r^2 Therefore, the electric flux through spherical surface of radius R will be   \phi=E_R 4 \pi R^2 When  r=R, E_R=\alpha R , we have  \phi = \alpha R4 \pi R^2 By Gauss theorem, net electric flux is  \dfrac{1}{\varepsilon_0} 	imes 	ext{change enclosed}  	herefore \alpha R 4 \pi R^2=\dfrac{1}{\varepsilon_0}Q_{enclosed} or   Q_{enclosed}=(4 \pi\varepsilon_0)\alpha R^3 Given R=0.30 m,  \alpha=100 Vm^{-2}  Q_{enclosed}=\dfrac{1}{9 	imes 10 ^9}	imes 100 	imes (0.30)^3=3 	imes 10^{-10}C

Question 10

Define electric flux. Apply Gauss' law to obtain an expression for the electric field intensity at a point due to an infinitely long uniformly charged straight wire. Draw the necessary diagram.

Accepted Answer

Electric flux is the measure of flow of the electric field through a given area. It is proportional to the number of electric field lines going through a normally perpendicular surface.For electric field  \vec{E}  passing through an area  \vec{A} , the flux passing through it is  \vec{E}.\vec{A} .According to Gauss' law,  flux passing through a closed surface is proportional to the amount of charge in the enclosed volume. \vec{E}.\vec{A}=\dfrac{q_{enc}}{\epsilon_0} Let us take a cylindrical surface co-axial with the charge carrying straight wire as Gaussian surface.Enclosed charge in this volume as shown in figure, assuming linear charge density to be  \lambda , is  q_{enc}=\lambda L Hence according to the Gauss' law, E(2\pi RL)=\dfrac{\lambda L}{\epsilon_0} \implies E=\dfrac{\lambda}{2\pi \epsilon_0 R}

Question 11

(a) Define electric flux. Is it a scalar or a vector quantity?
A point charge

q

is at a distance of

d / 2

directly above the centre of a square of side

d

, as shown in the figure. Use Gauss' law to obtain the expression for the electric flux through the square.
(b) If the point charge is now moved to a distance

^{'} d^{'}

from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.

Accepted Answer

\large 	extbf{Part (a)} 	extbf{Step 1: Definition of electric flux} 	extbf{Qualitative Definition:}  The Electric flux is proportional to the number of field lines passing a given area in a unit of time. This definition cannot be used for calculating the exact value of flux, it is used only for the comparison of flux through two surfaces. 	extbf{Quantitative Definition: }  Magnitude of Electric flux is given by                        \Delta \phi = \int \vec{E}.\vec{dS}                      For plane surface and uniform Electric field, the above equation becomes:                        \Delta \phi = \vec{E}.\vec{S}= ES\cos	heta It is a scalar quantity. 	extbf{Step 2: Electric flux through square } Suppose a cube enclosing the charge q of side  d . Hence according to Gauss's law, the electric flux through the cube is  \dfrac{q}{\epsilon_o} The given case can be considered as one of the face of such a cube.By symmetry, the flux through the face is  \dfrac{1}{6}th  of the flux through the cube.Hence flux through the square face =  \dfrac{q}{6 \epsilon_o} \large 	extbf{Part (b)} 	extbf{Electric flux through square when the side is doubled} Now, the charge is placed at a distance  d  from the centre of square. Also, the side of the square is  2d  in this case.In this case, the charge will be at the center of a cube of side 2d and the square will be one of its face.According to a similar analysis as shown in Step 2 the electric flux through the face is:  \dfrac{q}{6 \epsilon_o} Hence there will be no change in the electric flux.

Question 12

If the electric field at the centre of the semicircular ring shown in figure. is  \cfrac { -xkq\hat { i }  }{ \pi { R }^{ 2 } }  , then find  x .

Accepted Answer

Consider an element of length  Rd	heta  and charge on the element  \displaystyle dq=\lambda Rd	heta   where  \lambda=  charge per unit length.Here sin component of net electric filed due to element of both positive and negative charge will cancel each other and only cosine component contribute the filed. i.e, \displaystyle dE_O=2dE\cos	heta=2\frac{kdq}{R^2}\cos	heta=2\frac{k \lambda Rd	heta}{R^2}\cos	heta \displaystyle E=\frac{2k\lambda R}{R^2}\int_0^{\frac{\pi}{2}}\cos	heta d	heta=\frac{2k\lambda R}{R^2} Since  \displaystyle q=\lambda (\frac{\pi}{2})R,   thus  \displaystyle E=\frac{2k(2q)}{\pi R^2}=\frac{4qk}{\pi R^2} As net field in the direction of negative x axis so  \displaystyle \vec{E}=\frac{4qk}{\pi R^2} (-\hat i) thus,  x=4

Electric Charges And Fields