Electrostatics

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point ….

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Q: Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting , mass less string  of same length . At equilibrium the angle between the strings is θ . The spheres are now immersed  in a dielectric liquid of density 800 kg/m^3 and dielectric constant 21 . If the angle between the strings remains the same after the immersion then

(a) electric force between the spheres remains unchanged

(b) electric force between the spheres reduces

(c) mass density of the spheres is 840 kg/m^3

(d) the tension in the strings holding the spheres remains unchanged

Click to See Solution :
Ans: (b,c)

Sol: In air ,

T cosθ = m g …(i)

T sinθ = q E  …(ii)

on dividing , qE = m g tanθ
IIT
In liquid ,

T’cosθ = m g – FB

$T’ cos\theta = m g(1+\frac{\rho_l}{\rho_s})$ …(iii)

$T’ sin\theta = \frac{qE}{K}$ …(iv)

So , $\frac{T’}{T} = \frac{1}{K}$

$T’ = \frac{T}{K} = \frac{T}{21}$

On dividing (iv) by (iii)

$\frac{qE}{K} = m g (1-\frac{\rho_l}{\rho_s})tan\theta $

$qE = K m g (1-\frac{\rho_l}{\rho_s})tan\theta$

Solving , ρs = 840 kg/m^3

 

A point charge q of mass m is suspended vertically by a string of length l . A point dipole of dipole moment p is ….

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Q: A point charge q of mass m is suspended vertically by a string of length l . A point dipole of dipole moment p is now brought towards q from infinity so that charge moves away . The final equilibrium position of the system including the direction of the dipole , the angles and distances is shown in the figure below . If the work done in bringing the dipole to this position is N x (mgh) , where g is the acceleration due to gravity , then value of N is ——– (Note that for three coplanar forces keeping a point mass in equilibrium ,F/sinθ is the same for all forces , where F is any one of the forces and θ is the angle between the other two forces)

IIT

Click to See Solution :
Ans: (2)

Sol: IIT

Ui = 0

$\displaystyle U_f = \frac{k q p}{(2 l sin\frac{\alpha}{2})^2} + m gh $ ….(i)

From Triangle , α + 90°-θ + 90°-θ = 180°

α = 2θ
Since , $h = 2lsin(\frac{\alpha}{2})sin\theta $

$h = 2lsin(\frac{\alpha}{2})sin(\frac{\alpha}{2}) $

$h = 2lsin^2(\frac{\alpha}{2}) $

Now , charge is in equilibrium ,

$\frac{mg}{sin(90+\frac{\alpha}{2})} = \frac{qE}{sin(180-2\theta)}$

$\frac{mg}{cos\frac{\alpha}{2}} = \frac{qE}{sin2\theta}$

$\frac{mg}{cos\frac{\alpha}{2}} = \frac{qE}{sin\alpha}$

$\frac{mg}{cos\frac{\alpha}{2}} = \frac{qE}{2sin(\alpha/2)cos(\alpha/2)}$

$qE = mg 2 sin\frac{\alpha}{2}$

$q \frac{2kp}{(2lsin\frac{\alpha}{2})^3} = mg 2 sin\frac{\alpha}{2}$

$ \frac{qkp}{(2lsin\frac{\alpha}{2})^2} = mg sin\frac{\alpha}{2} \times 2lsin\frac{\alpha}{2}$

$ \frac{qkp}{(2lsin\frac{\alpha}{2})^2} = m g h $

Putting this value in eq^n (i)

$U_f = mgh + mgh = 2 m g h$

$W = \Delta U = N m g h $

N = 2

 

Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure …..

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Q: Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure . Charged oil drops of density 900 kg/m^3 are released through a tiny hole at the center of the top disc . Once some oil drops achieve terminal velocity , the switch is closed to apply a voltage of 200 V across the discs . As a result , an oil drop of radius 8 × 10-7 m stops moving vertically and floats between the discs . The number of electrons presents in this oil drop is ——– (neglect the buoyancy force , take acceleration due to gravity = 10 m/s^2 and charge on electron = e = 1.6 x 10-19 C)

IIT

Click to See Solution :
Ans: (6)
Sol: q E = m g
$\displaystyle q = \frac{m g}{E} = \frac{m g}{V/d}$

$\displaystyle q= \frac{\frac{4}{3}\pi (8\times 10^{-7})^3 \times 900 \times 10}{200/0.01}$

$\displaystyle N = \frac{q}{e} = \frac{\frac{4}{3}\pi (8\times 10^{-7})^3 \times 900 \times 10 \times 0.01}{200 \times 1.6 \times 10^{-19}} $

N = 6

 

A circular disc of radius R carries surface charge density σ(r) = σo(1-r/R) , where σo is constant r is the distance …..

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Q: A circular disc of radius R carries surface charge density σ(r) = σo(1-r/R) , where σo is constant r is the distance from the center of the disc . Electric flux through a large spherical surface that encloses the charged disc completely is φo . Electric flux through another spherical surface of radius R/4 and concentric with disc is φ . Then φo/φ is ———

Click to See Solution :
Ans: (6.40)

Sol:
IIT

$\displaystyle \phi_o = \frac{\int dq}{\epsilon_o}$

$\displaystyle \phi_o = \frac{1}{\epsilon_o} \int_{0}^{R} \sigma_o (1-\frac{r}{R})2\pi r dr $

$\displaystyle \phi_o = \frac{\sigma_o 2 \pi}{\epsilon_o} \int_{0}^{R} (r-\frac{r^2}{R}) dr $

$\displaystyle \phi = \frac{\sigma_o 2 \pi}{\epsilon_o} \int_{0}^{R/4} (r-\frac{r^2}{R}) dr $

On dividing ,

$\displaystyle \frac{\phi_o}{\phi} = \frac{\int_{0}^{R} (r-\frac{r^2}{R}) dr}{\int_{0}^{R/4} (r-\frac{r^2}{R}) dr} $

$\displaystyle \frac{\phi_o}{\phi} = \frac{R^2/2 – R^2/3}{R^2/32 – R^2/(3\times 64)} $

$\displaystyle \frac{\phi_o}{\phi} = \frac{32}{5} = 6.40 $