Question

two spheres of radii 5cm and 3cm are given charges on risk volume and 50 calling respectively and then connected by a wire calculate the loss of energy after connection​

Answer 1

Answer:

Solution given:

Radius of small sphere[r]=5cm=0.05m

Radius of large sphere[R]=10cm=0.1m

capacitance of small sphere[c]=4πε0r

=$4π*8.85×10^{-12}×0.05=5.56*10^{-12}F$

Charge for small sphere[Q1]=100C

Charge for small sphere[Q2]=50C

Potential difference [V1]=$\frac{charge}{capacitance}=\frac{100}{5.56*10^{-12}}=1.8×10^{13}$V

.

again

capacitance of small sphere[C]=4πε0R

=$4π*8.85×10^{-12}×0.1=1.11*10^{-11}F$

Potential difference [V2]=$\frac{charge}{capacitance}=\frac{50}{1.11*10^{-11}}=4.5×10^{12}$V

Now

Loss of energy:

$\frac{cC(V1-V2)^{2}}{2(c+C)}$

=$\frac{5.56*10^{-12}*1.11*10^{-11}(1.8*10^{13}-4.5*10^{12})^{2}}{2(5.56*10^{-12}+1.11*10^{-11})}$

=25Joule