Q: Find the charge in coulomb on 1\ gram\ ion of N^{3-} .

Charge on one ion of N^{3-} = 3\times 1.6\times 10^{-19} coluomb One g-ion = 6.02\times 10^{23} ions Thus, charge on one g-ion of N^{3-} = 3\times 1.6\times 10^{-19}\times 6.02\times 10^{23} = 2.89\times 10^{5} coulomb .

Q: On removing 2.9 \times 10^{19} electrons from a metal sheet,then the charge on metal sheet is

Given: n=2.9\times 10^{19} electrons e=1.6\times 10^{-19}C Since the electrons are being removed hence the metal sheet is gaining positive charge.Total charge =2.9\times 1.6\times 10^{19-19}C =+4.64C or \boxed{+5C\,\, \text{approx}} Option C

Q: Find force on semicircular ring due to point charge q .

Consider an element at an angle \theta charge dq=\lambda Rd\theta \therefore Force on charge dq, df=\dfrac{Kqdq}{R^{2}} =\dfrac{Kq\lambda Rd\theta}{R^{2}} Due to symmetry of the system we see F_{x} gets cancelled \therefore \displaystyle \int df_{y}=\dfrac{Kg\lambda}{R} \int_{0}^{\pi}\sin\theta d\theta =\dfrac{Kq\lambda}{R} \left[-\cos\theta \right]^{\pi}_{0} F=\dfrac{2Kq\lambda}{R}

Q: Electric charges q, q and -2q are placed at the corners of an equilateral triangle ABC of side L . The magnitude of electric dipole moment of the system is

Electric dipole is a vector quantity \therefore Components along X-axis of \overset{\rightarrow}P1 and \overset{\rightarrow}P2 cancel out each other(opposite direction). \therefore Components along Y-axis will add up as: \mid \overset{\rightarrow}P_{ret} \mid=[(q\,\cdot\,L)\cos 30^{\circ}]\times 2 =q\,\cdot\,L\,\cdot\cfrac{\sqrt 3}{2}\times 2=\sqrt 3qL

Q: A sphere of radius r has a volume density of charge \rho=kr . Find the electric field intensity at the surface of the sphere.

Given : Charge per unit volume of sphere \rho = kr Total charge enclosed by the sphere Q_{enc} = \int_o^R \rho (4\pi r^2) dr \therefore Q_{enc} = 4\pi k \int_o^R r^4 dr OR Q_{enc} = 4\pi k \times \dfrac{r^4}{4}\bigg|^{R}_o OR Q_{enc} = 4\pi k \times \dfrac{R^4 - 0}{4} \implies Q_{enc} = \pi k R^4 Let the surface of the sphere is the Gaussian surface.Using Gauss's law, \int E.dS = \dfrac{Q_{enc}}{\epsilon_o} \therefore E_{surface}\int dS = \dfrac{Q_{enc}}{\epsilon_o} OR E_{surface}\times 4\pi R^2 = \dfrac{\pi k R^4}{\epsilon_o} \implies E_{surface} =\dfrac{kR^2}{4\epsilon_o}

Question 1

Find the charge in coulomb on  1\ gram\ ion  of  N^{3-} .

Accepted Answer

Charge on one ion of  N^{3-} = 3	imes 1.6	imes 10^{-19} coluomb One  g-ion = 6.02	imes 10^{23} ions Thus, charge on one g-ion of  N^{3-} = 3	imes 1.6	imes 10^{-19}	imes 6.02	imes 10^{23} = 2.89	imes 10^{5} coulomb .

Question 2

On removing  2.9 	imes 10^{19}   electrons from a metal sheet,then the charge on metal sheet is

Accepted Answer

Given: n=2.9	imes 10^{19}  electrons e=1.6	imes 10^{-19}C Since the electrons are being removed hence the metal sheet is gaining positive charge.Total charge  =2.9	imes 1.6	imes 10^{19-19}C                         =+4.64C  or                   \boxed{+5C\,\, 	ext{approx}} Option  C

Question 3

Find force on semicircular ring due to point charge  q .

Accepted Answer

Consider an element at an angle  	heta  charge  dq=\lambda Rd	heta 	herefore  Force on charge  dq, df=\dfrac{Kqdq}{R^{2}} =\dfrac{Kq\lambda Rd	heta}{R^{2}} Due to symmetry of the system we see  F_{x}  gets cancelled  	herefore \displaystyle \int df_{y}=\dfrac{Kg\lambda}{R} \int_{0}^{\pi}\sin	heta d	heta                   =\dfrac{Kq\lambda}{R} \left[-\cos	heta ight]^{\pi}_{0}               F=\dfrac{2Kq\lambda}{R}

Question 4

Why do the electric field lines never cross each other?

Accepted Answer

At any point, if electric field lines cross each other it means at that point there are two directions of electric field, which is impossible.

Question 5

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 6

Electric charges  q, q  and  -2q  are placed at the corners of an equilateral triangle ABC of side  L . The magnitude of electric dipole moment of the system is

Accepted Answer

Electric dipole is a vector quantity 	herefore  Components along X-axis of  \overset{ightarrow}P1  and  \overset{ightarrow}P2  cancel out each other(opposite direction). 	herefore  Components along Y-axis will add up as: \mid \overset{ightarrow}P_{ret} \mid=[(q\,\cdot\,L)\cos 30^{\circ}]	imes 2 =q\,\cdot\,L\,\cdot\cfrac{\sqrt 3}{2}	imes 2=\sqrt 3qL

Question 7

Which of the following statements are correct?

Accepted Answer

According to Gauss's law,  (\phi=\dfrac{Q_{en}}{\epsilon_0})  the electric flux only depends on the charge enclosed by the surface. Gauss's law is applicable on any other distribution of charges but it is applicable only for symmetrical structure of the body.Ans: (C)

Question 8

A sphere of radius  r  has a volume density of charge  \rho=kr . Find the electric field intensity at the surface of the sphere.

Accepted Answer

Given : Charge per unit volume of sphere   ho = kr Total charge enclosed by the sphere   Q_{enc} = \int_o^R ho (4\pi r^2) dr 	herefore    Q_{enc} = 4\pi k \int_o^R r^4 dr OR    Q_{enc} = 4\pi k 	imes \dfrac{r^4}{4}\bigg|^{R}_o OR    Q_{enc} = 4\pi k 	imes \dfrac{R^4 - 0}{4}               \implies Q_{enc} = \pi k R^4 Let the surface of the sphere is the Gaussian surface.Using Gauss's law,   \int E.dS = \dfrac{Q_{enc}}{\epsilon_o} 	herefore     E_{surface}\int dS = \dfrac{Q_{enc}}{\epsilon_o} OR     E_{surface}	imes 4\pi R^2 = \dfrac{\pi k R^4}{\epsilon_o}          \implies E_{surface} =\dfrac{kR^2}{4\epsilon_o}

Question 9

A small conducting sphere carrying charge  Q  is surrounded by a spherical uncharged conducting shell. A small charge  +q  is placed between the sphere and the shell, then the net charge on the inner surface of the shell is :

Accepted Answer

For conducting sphere all charges are distributed over the surface. So charge Q will appear on outer surface of the conducting sphere. As the charge inside shell is zero when q is placed between the conducting sphere and shell ,  -(Q+q)  will appear on the inner surface of the shell.

Electric Charges And Fields

Quantum nature of charge

Electric field due to a charged semicircle

Electric field lines

Electric field due to a dipole

Electric flux through a closed surface in uniform electric field

Electric field due to a conducting cylinder

Electric field due to a uniformly charged thin spherical shell