Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

Two small equally charged spheres each of mass

m

, are suspended from the same point by silk threads of length

l

. The distance between the spheres

x ≪ l

. Find the rate

\frac{d q}{d t}

with which the charge leaks off each sphere, if their approach velocity varies as

v = \frac{a}{\sqrt{x}}

, where

a

is a constant.

Accepted Answer

Let us assume that the threads makes an angle $θ$ with the vertical, when the spheres are at a distance $x$ from each other. Since, the charge on each sphere is leaking, the distance between the spheres keeps on decreasing and the rate of change of this distance is given.

We know that the spheres are moving towards each other with some variable velocity. But the rate at which the charge leaks from any of the spheres is very small. Due to this, the approach velocity as well as the rate of change of approach velocity of the two spheres will be very small.

Hence, we can consider the two spheres to be momentarily in equilibrium. Applying Newton's second law of motion on any one of the spheres, we obtain,

$T cosθ = m g$ (vertical force balance)

$T sinθ = F_{e}$ (horizontal force balance)

From the above two equations, we get,

$tanθ = \frac{F_{e}}{m g}$

As it is clear from the figure, for a very small angle $θ$ ,

$tanθ = \frac{x}{2 \sqrt{l^{2} - {(\frac{x}{2})}^{2}}} \approx \frac{x}{2 l}$ , as $x ≪ l$

$\Rightarrow F_{e} = \frac{m g x}{2 l}$

$\Rightarrow \frac{q^{2}}{4 π ε_{0} x^{2}} = \frac{m g x}{2 l}$

$\Rightarrow q^{2} = \frac{2 π ε_{0} m g x^{3}}{l}$

$\Rightarrow q = \sqrt{\frac{2 π ε_{0} m g x^{3}}{l}}$

To find the rate at which the charge leaks off from the sphere, we have to differentiate charge with respect to time.

$\frac{d q}{d t} = \sqrt{\frac{2 π ε_{0} m g}{l}} (\frac{3}{2} \sqrt{x}) (\frac{d x}{d t})$

According to the problem, $\frac{d x}{d t} = v = \frac{a}{\sqrt{x}}$ (Approach velocity is $\frac{d x}{d t}$ )

Hence, $\frac{d q}{d t} = \frac{3 a}{2} \sqrt{\frac{2 π ε_{0} m g}{l}}$

Important Questions on ELECTRODYNAMICS

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Two small equally charged spheres each of mass m, are suspended from the same point by silk threads of length l. The distance between the spheres x≪l. Find the rate dqdt with which the charge leaks off each sphere, if their approach velocity varies as v=ax, where a is a constant.

Important Questions on ELECTRODYNAMICS

Learn and Prepare for any exam you want !