Answer:
Capacitance, [tex]C=2\ pF[/tex]
Explanation:
It is given that,
Charge stored in the capacitor, [tex]q=2\times 10^{-10}\ C[/tex]
Potential difference, V = 100 V
Capacitance of a system is given by :
[tex]C=\dfrac{q}{V}[/tex]
[tex]C=\dfrac{2\times 10^{-10}\ C}{100\ V}[/tex]
[tex]C=2\times 10^{-12}\ Farad[/tex]
or C = 2 picofarad , c = 12 pF
Hence, the capacitance of the system is [tex]2\ pF[/tex]
