Q: (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.

\large \textbf{Part (a)} \textbf{Step 1: Definition of electric flux} \textbf{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. \textbf{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\theta It is a scalar quantity. \textbf{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 \textbf{Part (b)} \textbf{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.

Q: A particle of charge 2 \mu C and mass 1.6 g is moving with a velocity 4 \hat{i} ms^{-1} . At t = 0 the particle enters in a region having an electric field \vec{E} (in NC^{-1}) = 80 \hat{i} + 60 \hat{j} . Find the velocity of the particle at t = 5 s .

Given that :-q = charge = 2 \mu C = 2 \times 10^{-6} C mass = m = 1.6 g = 1.6 \times 10^{-3} kg V = velocity = 4 \hat{i} m/s \vec{E} = (80 \hat{i} + 60 j ) N/C Since particle has initial velocity only in x-direction u_x = v = 4 m/s ; \, u_y = 0 m/s Force = F = q \vec{E} \Rightarrow \vec{F} = 2 \times 10^{-6} (80 \hat{i} + 60 j) = (160 \hat{i} + 120 j) \times 10^{-6} \vec{a} = acceleration = \vec{a} = \dfrac{\vec{F}}{m} = \dfrac{(160 \hat{i} + 120 j) \times 10^{-6}}{1.6 \times 10^{-3}} \vec{a} = (100 \hat{i} + 75 \hat{j}) \times 10^{-3} = (a_x \hat{i} + a_y \hat{j}) time = t = 5 sec v_x = u_x + a_x t = (4 + 100 \times 10^{-3} \times 5) v_x = 4 + 0.5 = 4.5 m/s v_y = u_y + a_y t = 0 + 75 \times 10^{-3} \times 5 = 0.375 m/s Where u_x , v_x are initial and final velocity in x-direction and u_y and v_y are initial and final velocity in y-direction. \therefore Final velocity = v = v_x \hat{i} + v_y \hat{j} = 4.5 \hat{i} + 0.375 j v = (4.5 \hat{i} + 0.375 \hat{j}) m/s

Question 1

Derive an expression for the electric field E due to a dipole of length  '2a'  at a point distant r from the centre of the dipole on the axial line.

Accepted Answer

Electric field at point P at a distance r from centre of dipole due to +q E_B=\frac{1}{4\pi\epsilon_0}\frac{q}{BP^2}  along BP    = \frac{1}{4\pi\epsilon_0}\frac{q}{(r-a)^2} Electric field at point P due to -q is E_A=\frac{1}{4\pi\epsilon_0}\frac{q}{AP^2}  along PA    = \frac{1}{4\pi\epsilon_0}\frac{q}{(r+a)^2} Net field at P is given by  E_P = E_B-E_A = \frac{1}{4\pi\epsilon_0}(\frac{q}{(r-a)^2}-\frac{q}{(r+a)^2})=\frac{q}{4\pi\epsilon_0}\frac{4ar}{(r^2-a^2)^2}=\frac{1}{4\pi\epsilon_0}\frac{2pr}{(r^2-a^2)^2} &#10; along BP . Where p is the dipole=2qa

Question 2

An infinitely large thin plane sheet has a uniform surface charge density  +\sigma . Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet.

Accepted Answer

Let PQRS is an infinitely plane charged sheet and  +\sigma  be the distribution of charge over the sheet.Let the charge  q  is bringing from infinity to the distance  r  from the sheet.Then, Electric field at the distant r from the sheet \phi=E.	riangle S cos	heta 	herefore \phi=E.	riangle S (	heta=0) ----------(1)and  \phi=\dfrac{\sigma}{\epsilon_o} -------------(2)From equation (1) and (2) E=\dfrac{\sigma}{\epsilon_o	imes	riangle S} E=\dfrac{\sigma}{\epsilon_o	imes2} We know that, W=F.r W= q.\dfrac{\sigma}{\epsilon_o	imes 2}.r

Question 3

Fill in the blank with appropriate answer :Electric flux through a spherical surface shown in the figure, is _______ .

Accepted Answer

Electric flux through the surface =  \dfrac{q_{in}}{ \varepsilon_0} from the given figure, q_{in} = +q_2 - q_1 	herefore \phi = \dfrac{q_{in}}{ \varepsilon_0} = \left(\dfrac{q_2 - q_1}{ \varepsilon_0} ight)

Question 4

(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 5

Five point charges, each of value +q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value -q placed at the centre of the hexagon?

Accepted Answer

Among the five forces on five charges,2pairs(four) of forces are cancelled with each other since they are equal in magnitude and opposite in direction.The resultant force is due to a single force given below. F=\dfrac{1}{4\pi\epsilon_{\circ}}	imes \dfrac{q	imes q}{l^2} F=\dfrac{1}{4\pi\epsilon_{\circ}}	imes \dfrac{q^2}{l^2} (attractive in nature)

Question 6

Two metallic spheres  A  and  B  kept on insulating stands are in contact with each other. A positively charged rod  P  is brought near the sphere  A  as shown in the figure. The two spheres are separated from each other, and the rod  P  is removed. What will be the nature of charges on spheres  A  and  B ?

Accepted Answer

This process is called induction of charge and happens instantly.Separate the spheres by a small distance while the glass rod is still near sphere  A . The two spheres will be oppositely charged and attract each other and after remaining rod they will rearrange each other.

Question 7

A metal sphere is kept on an insulating stand. A negatively charged rod is brought near it, then the sphere is earthed as shown. On removing the earthing, and taking the negatively charged rod away, what will be the nature of charge on the sphere? Give reason for your answer.

Accepted Answer

A rod is brought close to the sphere, the free electron in the sphere move away due to deficit of electrons. The process of change distribution stops when the net force on the free electron inside the metal is less.When we connect the sphere to the ground electron will fow to the ground while the charge will remain on the sphere.So final charge on the sphere will be positive.

Question 8

Explain briefly, using a paper diagram, the difference in behaviour of a conductor and a dielectric in the presence of external electric field.

Accepted Answer

(a) When a conductor is placed in an electric field, charge distribution takes place. This distribution  in such a way that the electric field is zero inside the conductor (that is after the charge is distributed and reached a stable state)The electric field due to the induced charge concels the external (Ref image 1.)When a dielectric is placed in the external electric field, an induced electric field appears in the dielectric in the opposite direction of external electric field, [having the magnitude  \frac{E_0}{k}  as net field(Refer image2.)

Question 9

A particle of charge  2 \mu C  and mass  1.6 g  is moving with a velocity  4 \hat{i} ms^{-1}  . At  t = 0  the particle enters in a region having an electric field  \vec{E}  (in  NC^{-1}) = 80 \hat{i} + 60 \hat{j} . Find the velocity of the particle at  t = 5 s .

Accepted Answer

Given that :-q = charge  = 2 \mu C = 2 	imes 10^{-6} C mass  = m = 1.6 g = 1.6 	imes 10^{-3} kg V =   velocity  = 4 \hat{i} m/s \vec{E} = (80 \hat{i} + 60 j ) N/C Since particle has initial velocity only in x-direction u_x = v = 4 m/s ; \, u_y = 0 m/s Force  = F = q \vec{E} \Rightarrow \vec{F} = 2 	imes 10^{-6} (80 \hat{i}  + 60 j) = (160 \hat{i} + 120 j) 	imes 10^{-6} \vec{a} =   acceleration  = \vec{a} = \dfrac{\vec{F}}{m} = \dfrac{(160 \hat{i} + 120 j) 	imes 10^{-6}}{1.6 	imes 10^{-3}} \vec{a} = (100 \hat{i} + 75 \hat{j}) 	imes 10^{-3} = (a_x \hat{i} + a_y \hat{j}) time  = t = 5 sec v_x = u_x + a_x t = (4 + 100 	imes 10^{-3} 	imes 5) v_x = 4 + 0.5 = 4.5 m/s v_y = u_y + a_y t = 0 + 75 	imes 10^{-3} 	imes 5 = 0.375 m/s Where  u_x , v_x  are initial and final velocity in x-direction and  u_y  and  v_y  are initial and final velocity in y-direction. 	herefore   Final velocity  = v = v_x \hat{i} + v_y \hat{j} = 4.5 \hat{i} + 0.375 j v = (4.5 \hat{i} + 0.375 \hat{j}) m/s

Electric Charges And Fields