Category: Electrostatics

Q: Suppose the charge of a proton and an electron differ slightly . One of them is -e , the other is (e + Δe) . If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size ) apart is zero , then Δe is of the order of [given mass of hydrogen m_h = 1.67 × 10^-27 kg]

(a) 10^-23 C

(b) 10^-37 C

(c) 10^-47 C

(d) 10^-20 C

Ans: (b)

Sol: $F_g = F_e $

$ \frac{G m^2}{r^2} = \frac{1}{4\pi \epsilon_0}\frac{(\Delta e)^2}{r^2} $

$ \Delta e = \sqrt{G m^2 4 \pi \epsilon_0}$

$ = \sqrt{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2 \times \frac{1}{9 \times 10^9} } $

= 10^-37 C

Q: An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field .The direction of electric field is now reversed , keeping its magnitude same . A proton is allowed to fall from rest in it through same vertical distance h . The time of fall of electron , in comparison to the time of fall of proton is

(a) 10 times greater

(b) Equal

(c) 5 times greater

(d) Smaller

Ans: (d)

Sol: Acceleration $ \displaystyle a = \frac{F}{m} = \frac{e E}{m}$

$ \displaystyle h = 0 + \frac{1}{2} a t^2 $

$ \displaystyle h = \frac{1}{2} \frac{e E}{m} t^2 $

$ \displaystyle t^2 = \frac{2 m h}{e E} $

$ \displaystyle t \propto \sqrt{m} $

As , m_e < m_p

t_e < t_p

A toy car with charge q moves on a friction less horizontal plane surface under the influence of a uniform electric field $ \vec{E}$ . Due to the force $\vec{q E}$ , its velocity increases from 0 to 6 m/s in one second duration . At that instant the direction of field is reversed . The car continue to move for two more seconds under the influence of this field . The average velocity and average speed of the toy car between 0 to 3 seconds are respectively .

(a) 1 m/s , 3.5 m/s

(b) 1.5 m/s , 3 m/s

(c) 1 m/s , 3 m/s

(d) 2 m/s , 4 m/s

Sol: velocity increases from 0 to 6 m/s in one second

Acceleration , $ \displaystyle a = \frac{6-0}{1} = 6 m/s^2 $

$\displaystyle S_1 = \frac{1}{2}a (1)^2 $

$ \displaystyle S_1 = \frac{1}{2}\times 6 \times (1)^2 = 3 m$

Now , the direction of field is reversed , hence motion being retarded & momentarily comes to rest after next one second , after that direction is reversed .

$\displaystyle S_2 = 6 \times 1 – \frac{1}{2}\times 6 \times (1)^2 $

$ \displaystyle S_2 = 3 m $

$\displaystyle S_3 = \frac{1}{2}a (1)^2 = 3 m$

Average speed$\displaystyle = \frac{Total \;Distance}{Total \; time} $

$\displaystyle = \frac{9}{3} = 3 m/s$

Average velocity $\displaystyle = \frac{Displacement}{time} $

$ \displaystyle = \frac{3}{3} = 1 m/s$

Q: The electrostatic potential inside a charged spherical ball is given by φ = a r² + b where r is the distance from the centre ; a , b are constants. Then charge density inside ball is

(a) -6 a ε₀ r

(b) -24 π a ε₀ r

(c) -6 a ε₀

(d) -24 π a ε₀

Ans:(c)

Sol: Potential inside a charged spherical ball

φ = a r² + b

$ \displaystyle E = -\frac{d\phi}{dr} = -2 a r $

According to Gauss’s theorem ,

E (4πr²) = q/ε₀

Differentiating ,

$ \displaystyle d(4\pi r^2 E) = \frac{dq}{\epsilon_0} = \frac{\rho (4\pi r^2 dr)}{\epsilon_0} $ ; Where ρ = volume charge density

$ \displaystyle r^2 dE + E(2r dr) = \frac{1}{\epsilon_0}\rho r^2 dr $

$ \displaystyle \frac{dE}{dr} + \frac{2E}{r} = \frac{\rho}{\epsilon_0} $

$ \displaystyle \frac{d}{dr}(-2ar) + \frac{2}{r}(-2ar) = \frac{\rho}{\epsilon_0} $

$ \displaystyle -2a – 4a = \frac{\rho}{\epsilon_0} $

$ \displaystyle – 6a = \frac{\rho}{\epsilon_0} $

ρ = -6 a ε₀

Q: A 4 μF capacitor and a resistance of 2.5 M Ω are in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given In (2) = 0.693]

(a) 13.86 s

(b) 6.93 s

(c) 7 s

(d) 14 s

Ans: (a)

Sol: During the growth of voltage in R – C circuit , the voltage across a capacitor at time t is given by

$\displaystyle V = V_0 (1- e^{-t/RC}) $

As , V_C = 3 V_R

and , V_C + V_R = V₀

V_C = (3/4)V₀

$ \displaystyle \frac{3}{4}V_0 = V_0 (1- e^{-t/RC}) $

$\displaystyle e^{-t/RC} = \frac{1}{4} $

$\displaystyle e^{t/RC} = 4 $

$ \displaystyle e^{t/RC} = 2^2 $

Taking log ,

$ \displaystyle \frac{t}{RC}log_e e = 2 log_e 2 $

$ \displaystyle t = 2 R C log_e 2 $

t = 2 × 2.5 × 10⁶ × 4 × 10^-6 × 0.693

t = 13.86 sec