Q: An electric dipole is formed by two particles fixed at the ends of a light rigid rod of length l . The mass of each particle is m and charges are -q and +q . The system is suspended by a torsionless thread in an electric field of intensity E such that the dipole axis is parallel to the field. If it is slightly displaced, the period of angular motion is :

We know torque on a dipole is \tau=P\times E When the dipole is displaced by \theta, \tau=PE sin\theta The electric field tries to align the dipole in the same direction as that of the field. \therefore \theta and \alpha (angular acceleration) are always in opposite directions.we know \tau=I\alpha I\alpha = -PE sin\theta sin\theta = \theta for small \theta \Rightarrow \alpha=\dfrac {-P E \theta}{I} we know in S. H. M , a=-\omega ^2 x \therefore \omega=\sqrt {\dfrac {PE}{I}} I at center of rod =\dfrac {m l^2}{2} \Rightarrow \omega=\sqrt {\dfrac {2 \cdot q\cdot l\cdot E }{m l^2}} \therefore T=\dfrac {2\pi}{\omega} and frequency f=\dfrac {1}{T}=\dfrac {1}{2\pi}\sqrt {\dfrac {2 q E}{m l}} T=2\pi \sqrt {\dfrac {m l}{2 q E}}

Question 1

Figure shows lines of force for a system of two point charges. The possible choice for the charges is:

Accepted Answer

As lines of forces are coming out of  q_1  and going into  q_2 q_1ightarrow  positive  charge q_2ightarrow  negative  chargeAs number of lines of forces are greater near  q_1 , then: \mid q_1\mid >\mid q_2\mid 	herefore  option A is correct.

Question 2

In a region where intensity of electric field is  5NC^{-1}  ,  40  lines of electric force are crossing per square metre. The number of lines crossing per square metre where intensity of electric field is  10NC^{-1}  will be :

Accepted Answer

Number of lines of electric force crossing is equal to flux  (\phi) We know  \phi =\int E.da 	o  By keeping area constant and doubling E,  \phi  will also get doubled. 	herefore  Number of lines get doubled  =80

Question 3

An electric dipole is formed by two particles fixed at the ends of a light rigid rod of length  l . The mass of each particle is m and charges are  -q  and  +q . The system is suspended by a torsionless thread in an electric field of intensity  E  such that the dipole axis is parallel to the field. If it is slightly displaced, the period of angular motion is :

Accepted Answer

We know torque on a dipole is  	au=P	imes E When the dipole is displaced by  	heta, 	au=PE sin	heta The electric field tries to align the dipole in the same direction as that of the field.  	herefore 	heta   and   \alpha  (angular acceleration) are always in opposite directions.we know  	au=I\alpha I\alpha = -PE sin	heta  sin	heta = 	heta  for small    	heta \Rightarrow \alpha=\dfrac {-P E 	heta}{I}  we know in S. H. M ,   a=-\omega ^2 x 	herefore \omega=\sqrt {\dfrac {PE}{I}} I at center of rod  =\dfrac {m l^2}{2} \Rightarrow \omega=\sqrt {\dfrac {2 \cdot q\cdot l\cdot E }{m l^2}} 	herefore T=\dfrac {2\pi}{\omega} and frequency  f=\dfrac {1}{T}=\dfrac {1}{2\pi}\sqrt {\dfrac {2 q E}{m l}} T=2\pi \sqrt {\dfrac {m l}{2 q E}}

Question 4

Charge on an originally uncharged conductor is separated by holding a positively charged rod very closely nearly, as in Fig. Assume that the induced negative charge on the conductor is equal to the positive charge  q  on the rod. Then, flux through surface  S_{1}  is

Accepted Answer

Initial charge on conductor  =0 Due to charge rod, negative and positive charge will induce at near and far point of the rod. 	herefore   Induced positive charge  = -(induced negative charge)                                                   =q On surface  S_1 , applying gauss' law : \displaystyle  \int_{s_1}  \overrightarrow{E}.\overrightarrow{ds}=\frac{Qenc}{E_o} 	herefore   Flux through  S_1 =\displaystyle \frac{Total\ \ charged \ \ enclosed}{E_o}                                         \displaystyle =\frac{q+(q-q)}{E_o}                                         \displaystyle =\frac{q}{E_o}

Question 5

An infinite, uniformly charged sheet with surface charge density  \sigma  cuts through a spherical Gaussian surface of radius R at a distance x from its center, as shown in the figure. The electric flux through the Gaussian surface is :

Accepted Answer

Applying Gauss's law, electric flux, \phi=\dfrac{Q_{enclosed}}{\epsilon_0} Here charge enclosed by Gaussian surface,  Q_{enclosed}=\pi(R^2-x^2)\sigma So, electric flux through the Gaussian surface,  \phi=\dfrac{\pi(R^2-x^2)\sigma}{\epsilon_0} Ans: (D)

Question 6

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 7

A point of charge  + Q  is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is :-

Accepted Answer

Our experts are building a solution for this

Electric Charges And Fields