Q: Two particles having charges { q }_{ 1 } and { q }_{ 2 } exert a force F on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?

F=k\dfrac{q_1q_2}{r^2} F'=k\dfrac{q'_1q'_2}{r'^2} given, q'_1=2q_1, q'_2=2q_2, r'=\dfrac{r}{2} \therefore \dfrac{F'}{F} =\dfrac{q'_1q'_2}{q_1q_2}\dfrac{r^2}{r'^2}=4\times 4=16 F'=16 F ans: (D)

Q: A long string with a charge of \lambda per unit length passes through an imaginary cube of edge length a . The maximum flux of the electric field through the cube will be:

\textbf{Hint} : Use the expression to find flux using the Gauss law. \textbf{Step 1:Total Electric Flux} :We can find charge using charge density ie..e q=\lambda l The flux passing through any surface enclosing charge q is \dfrac{q}{\epsilon_0} .Given a long string with a charge of \lambda per unit length passes through an imaginary cube of edge length a . \textbf{Step 2:Maximum Flux through Cube} To get maximum flux outside the cube, the maximum length of this string will have to pass through the cube. So, we choose a cross diagonal orientation of such a string i.e. body diagonal of cube. The total length of the string in the box is obtained using pythagoras theorem. l=\sqrt{2a^{2}+a^{2}} l=\sqrt(3)a Substitute l=\sqrt{3}a in q=\lambda l We get q=\lambda \sqrt(3)a We get maximum flux \phi=\dfrac{\lambda \sqrt(3)a}{\epsilon_0} .Thus option D is correct.

Question 1

Two charges each of  100  micro coulomb are separated in a medium of relative permittivity  2  by a distance of  5 cm . The force between them is :

Accepted Answer

We know force between two charges in a medium separated by a distance 'r' F=\dfrac {(q_1)(q_2)}{4\pi \varepsilon r^2} where,   \epsilon_r= \dfrac{\epsilon}{\epsilon_o} given,  \epsilon_r=2   \implies \epsilon = 2	imes \epsilon_o F=\dfrac {(100	imes 10^{-6})(100	imes 10^{-6})	imes 9	imes 10^9}{2	imes (5	imes 10^{-2})^2} F=\dfrac {90}{2	imes 25	imes 10^{-4}} F=\dfrac {90	imes 100	imes 10^2}{50} F=1\cdot 8	imes 10^4 N

Question 2

Two particles having charges  { q }_{ 1 }  and  { q }_{ 2 }  exert a force  F  on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10; F=k\dfrac{q_1q_2}{r^2} &#10;&#10; F'=k\dfrac{q'_1q'_2}{r'^2} &#10;&#10;given,  q'_1=2q_1,&#10;q'_2=2q_2, r'=\dfrac{r}{2} &#10;&#10; 	herefore \dfrac{F'}{F} =\dfrac{q'_1q'_2}{q_1q_2}\dfrac{r^2}{r'^2}=4	imes&#10;4=16 &#10;&#10; F'=16 F &#10;&#10;ans: (D)&#10;&#10;

Question 3

Four charges are arranged at the corners of a square ABCD as shown in figure. The force on a positive charge 'Q' kept at the centre of the square is :

Accepted Answer

The force on  +Q  at the center of square is, \vec{F}=k\left[\dfrac{-qQ(\vec{r_0}-\vec{r_1})}{|\vec{r_0}-\vec{r_1}|^3}+\dfrac{qQ(\vec{r_0}-\vec{r_2})}{|\vec{r_0}-\vec{r_2}|^3}+\dfrac{-2qQ(\vec{r_0}-\vec{r_3})}{|\vec{r_0}-\vec{r_3}|^3}+\dfrac{2qQ(\vec{r_0}-\vec{r_4})}{|\vec{r_0}-\vec{r_4}|^3}\right]  =\dfrac{kqQ}{r^2}\left[(\hat i+\hat j)-(-\hat i+\hat j)+2(-\hat i-\hat j)-2(\hat i-\hat j)\right] =\dfrac{kqQ}{r^2}\left[\hat i+\hat j+\hat i-\hat j-2\hat i-2\hat j-2\hat i+2\hat j\right] =\dfrac{kqQ}{r^2}(-2\hat i) thus,  F  is perpendicular to side AB.

Question 4

Question 5

The dipolemoment of the given system is:

Accepted Answer

See the diagram.Dipole moment of a system of charges is given by,  \overrightarrow p = \Sigma q_i \overrightarrow{r_i}  \vec P = -ql\hat{i} + ql\hat{i} + (-2q)(\sqrt{3}/2	imes l )\hat{j} = -\sqrt{3} ql \hat{j}

Question 6

A long string with a charge of  \lambda   per unit length passes through an imaginary cube of edge length  a . The maximum flux of the electric field through the cube will be:

Accepted Answer

extbf{Hint} : Use the expression to find flux using the Gauss law. 	extbf{Step 1:Total Electric Flux} :We can find charge using charge density ie..e  q=\lambda l The flux passing through any surface enclosing charge  q  is  \dfrac{q}{\epsilon_0} .Given a long string with a charge of  \lambda  per unit length passes through an imaginary cube of edge length  a .  	extbf{Step 2:Maximum Flux through Cube} To get maximum flux outside the cube, the maximum length of this string will have to pass through the cube. So, we choose a cross diagonal orientation &#10;of such a string i.e. body diagonal of cube. The total length of the string in the box is  obtained using pythagoras theorem.  l=\sqrt{2a^{2}+a^{2}} l=\sqrt(3)a Substitute  l=\sqrt{3}a  in   q=\lambda l We get  q=\lambda \sqrt(3)a We get maximum flux \phi=\dfrac{\lambda \sqrt(3)a}{\epsilon_0} .Thus option D is correct.

Question 7

A charged particle q of mass m is in equilibrium at a height h from a horizontal infinite line charge with uniform linear charge density  \lambda . The charge lies in the vertical plane containing the line charge. If the particle is displaced slightly (vertically), prove that the motion of the charged particle will be simple harmonic. Also find its time period

Accepted Answer

At equilibrium  mg = qE \displaystyle mg = \frac{q 2 k \lambda}{h}                 (i)When particle is displaced to a distance x in upward direction, then net force in upward direction is \displaystyle F = q \left ( \frac{2k \lambda}{h + x} ight ) - mg = \displaystyle \frac{2 k q \lambda}{h} \left ( 1 + \frac{x}{h} ight )^{-1} - mg = \frac{2 k q \lambda}{h} \left ( 1 - \frac{x}{h} ight ) - mg = \displaystyle - \left ( \frac{2 K q \lambda}{h^2} ight ) x = - K_0 x 	herefore F \propto - x   (Motion is SHM)Time period of motion is \displaystyle T = 2 \pi \sqrt{\frac{m}{K_0}} = 2 \pi \sqrt{\frac{mh^2}{2 kq \lambda}}

Question 8

The total solid angle subtended by the sphere is

Accepted Answer

Solid angle is given by area subtended divided by  r^2 . \Omega=\dfrac{\Delta S}{r^2} Now, area subtended by a sphere is the whole surface area of sphere= 4\pi r^2 \implies \Omega=\dfrac{4\pi r^2}{r^2} \implies \Omega=4\pi Answer-(E)

Question 9

Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

Accepted Answer

F=\dfrac{KQ^2}{r^2} If 25% of charge of A transferred to B then  q_A = Q - \dfrac{Q}{4} = \dfrac{3 Q}{4}   and  q_B = -Q + \dfrac{Q}{4} = \dfrac{-3Q}{4} F_1 = \dfrac{kq_A q_B}{r^2} F_1 = \dfrac{k \left(\dfrac{3Q}{4} \right)^2}{r^2} F_1 = \dfrac{9}{16} \dfrac{kQ}{r^2} \implies F_1 = \dfrac{9F}{16}

Question 10

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)

Electric Charges And Fields