Answer:
The energy lost is zero.
Explanation:
Given that,
First capacitor = 50 pF
Second capacitor = 200 pF
Potential = 4.2 kV
We need to calculate the energy lost
Using formula of energy lost
[tex]E = E_{initial}-E_{final}[/tex]
Put the value into the formula
[tex]E=\dfrac{1}{2}C_{1}V^2+\dfrac{1}{2}C_{2}V^2-\dfrac{}{}(C_{1}+C_{2})V^2[/tex]
[tex]E=\dfrac{1}{2}\times50\times10^{-12}\times(4.2\times10^{3})^2+\dfrac{1}{2}\times200\times10^{-12}\times(4.2\times10^{3})^2-\dfrac{1}{2}\times(50\times10^{-12}+200\times10^{-12})\times(4.2\times10^{3})^2[/tex]
[tex]E=0\ J[/tex]
Hence, The energy lost is zero.
