Category: Physics Q & A

Questions & Answers

The electrostatic potential inside a charged spherical ball is given by φ = a r^2 + b where r is the distance

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

A 4 μF capacitor and a resistance of 2.5 M Ω are in series with 12 V battery. Find the time after which…

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 , VC = 3 VR

and , VC + VR = V0

VC = (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

Two identical particles of charge q each are connected by a massless spring of force constant K…

Q: Two identical particles of charge q each are connected by a massless spring of force constant K. They are placed over a smooth horizontal surface, Fig. They are released when separation between them is r and the spring is unstretched. If maximum extension of the spring is r , then value of K ( neglecting gravitational effects) is

(a) \displaystyle \frac{q}{4 r}\sqrt{\frac{1}{\pi \epsilon_0 r}}

(b) \displaystyle \frac{q}{2 r}\sqrt{\frac{1}{\pi \epsilon_0 r}}

(c) \displaystyle \frac{2 q}{ r}\sqrt{\frac{1}{\pi \epsilon_0 r}}

(d) \displaystyle \frac{q}{ r}\sqrt{\frac{1}{\pi \epsilon_0 r}}

Ans: (b)

Sol: When the spring extends by r, elastic potential energy of spring \displaystyle = \frac{1}{2}K r^2

Change in P.E. of configuration of charges \displaystyle = \frac{1}{4 \pi \epsilon_0}(\frac{q^2}{r} - \frac{q^2}{r+r} )

\displaystyle = \frac{q^2}{8 \pi \epsilon_0 r}

Hence , \displaystyle \frac{1}{2}K r^2 = \frac{q^2}{8 \pi \epsilon_0 r}

\displaystyle K = \frac{q}{2 r}\sqrt{\frac{1}{\pi \epsilon_0 r}}

A solid conducting sphere having a charge Q is surrounded by an uncharged conducting hollow spherical shell…

Q: A solid conducting sphere having a charge Q is surrounded by an uncharged conducting hollow spherical shell. Let the pot. diff. between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge – 3 Q , the new potential difference between the same two surface is

(a) V

(b) 2 V

(c) 4 V

(d) – 2 V

Ans: (a)

Sol: Potential of inner sphere = Q/R1

Potential of outer surface of shell = Q/R2

Potential difference , \displaystyle V_1 = \frac{Q}{R_1} -\frac{Q}{R_2}

When a charge – 3 Q is given to the shell,

Potential of inner sphere , \displaystyle = \frac{Q}{R_1} -\frac{3Q}{R_2}

Potential of outer surface of shell , \displaystyle = \frac{Q}{R_2} -\frac{3Q}{R_2} = -\frac{2Q}{R_2}

Potential difference , \displaystyle V_2 = \frac{Q}{R_1} -\frac{3Q}{R_2} - (-\frac{2Q}{R_2})

\displaystyle V_2 = \frac{Q}{R_1} -\frac{Q}{R_2}

i.e. V2 = V1 = V

A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such that the surface densities are equal…

Q: A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such that the surface densities are equal. The potential at the common centre is

(a) zero

(b) Q(R+r)

(c) \displaystyle \frac{Q(R^2 + r^2)}{4 \pi \epsilon_0 (R + r)}

(d) \displaystyle \frac{Q(R + r)}{4 \pi \epsilon_0 (R^2 + r^2)}

Ans:(d):

Sol: Let q1 , q2 be charges on inner and outer spheres respectively.

\displaystyle \sigma_1 = \sigma_2 (Since , surface charge densities are equal)

\displaystyle \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2}

\displaystyle \frac{q_1}{q_2} = \frac{r^2}{R^2}

\displaystyle \frac{q_1}{q_2}+1 = \frac{r^2}{R^2}+1

\displaystyle \frac{q_1 + q_2}{q_2} = \frac{r^2 + R^2}{R^2}

\displaystyle \frac{Q}{q_2} = \frac{r^2 + R^2}{R^2}

\displaystyle q_2 = \frac{Q R^2}{R^2 + r^2}

Similarly , \displaystyle q_1 = \frac{Q r^2}{R^2 + r^2}

Potential at common centre is

\displaystyle V = \frac{q_1}{4\pi \epsilon_0 r } + \frac{q_2}{4\pi \epsilon_0 R}

\displaystyle = \frac{1}{4\pi \epsilon_0}[\frac{Q r}{R^2 + r^2} + \frac{QR}{R^2 + r^2}]

\displaystyle V = \frac{Q}{4\pi \epsilon_0} \frac{R+r}{R^2 + r^2}