Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

The figure shows two conducting spheres separated by a large distance and of radius $2 cm$ and $3 cm$ containing charges $10 μC$ and $20 μC$ , respectively. When the spheres are connected by a conducting wire, find out the following

$(i)$ The ratio of the final charges.

$(ii)$ Final charge on each sphere.

$(iii)$ The ratio of final charge densities.

(iv)

The heat produced during the process.

Accepted Answer

$(i)$ Let us consider that after connecting through the wire the spheres reach, common potential and their charges are ${q_{1}}^{'}$ and $10 μC$

$V_{c} = \frac{k {q_{1}}^{'}}{0.02} = \frac{k {q_{2}}^{'}}{0.03}$

$\Rightarrow \frac{{q_{1}}^{'}}{{q_{2}}^{'}} = \frac{2}{3}$ .

$(ii)$ From the conservation of the charge

$q_{1} + q_{2} = {q_{1}}^{'} + {q_{2}}^{'}$

$\Rightarrow 10 + 20 = \frac{2 {q_{2}}^{'}}{3} + {q_{2}}^{'}$

$\Rightarrow {q_{2}}^{'} = 18 μC$ and ${q_{1}}^{'} = 12 μC$ .

$(iii)$ The ratio of chagre densities, $\frac{σ_{1}}{σ_{2}} = \frac{\frac{{q_{1}}^{'}}{4 π {r_{1}}^{2}}}{\frac{{q_{2}}^{'}}{4 π {r_{2}}^{2}}} = \frac{{q_{1}}^{'} \times {r_{2}}^{2}}{{q_{2}}^{'} \times {r_{1}}^{2}}$

$\Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{12 \times 3^{2}}{18 \times 2^{2}}$

$\Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{3}{2}$ .

$(iv)$ Considering them isolated conductors having a capacitance $C = 4 π ε_{0} r$

$U_{i} = \frac{{q_{1}}^{2}}{2 C_{1}} + \frac{{q_{2}}^{2}}{2 C_{2}}$

$\Rightarrow \frac{{(10 \times 10^{- 6})}^{2}}{2 \times \frac{1}{(9 \times 10^{9} \times 0.02)}} + \frac{{(20 \times 10^{- 6})}^{2}}{2 \times \frac{1}{(9 \times 10^{9} \times 0.03)}}$

$\Rightarrow U_{i} = 22.5 + 60 = 82.5 J$

$U_{f} = \frac{{q_{1}}^{'}^{2}}{2 C_{1}} + \frac{{q_{2}}^{'}^{2}}{2 C_{2}}$

$\Rightarrow \frac{{(12 \times 10^{- 6})}^{2}}{2 \times \frac{1}{(9 \times 10^{9} \times 0.02)}} + \frac{{(18 \times 10^{- 6})}^{2}}{2 \times \frac{1}{(9 \times 10^{9} \times 0.03)}}$

$\Rightarrow U_{i} = 32.4 + 48.6 = 81.0 J$ .

Therefore, the heat developed $= U_{1} - U_{2} = 82.5 - 81.0 = 1.5 J$ .

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