Obtain the equivalent capacitance of the network in figure below. For a 300V supply determine the charge and voltage across each capacitor.

Explanation

`C_("eff")=(200 xx 100 xx 10^(-12) xx 10^(-12))/(10^(-12) (300))=(200)/(3)pF`
`Q=CV=(200)/(3) xx 10^(-12) xx 300 =2 xx 10^(-8)C`
`V_("100pF") (C_(4))=(2 xx 10^(-8))/(100 xx 10^(-12))=200V`
`V_("100pF") (C_(1))=100V`
`Q=100 xx 100^(-12)=10^(-8)C`
`Q_("200pF")=200 xx 10^(-12) xx 50=10^(-8)C`