Answer :

Capacitance of capacitor C1 = 100pF, C2 = 200pF, C3 = 200 pF, C4 = 100 pF.

Supply potential (V) = 300 V

Let the equivalent capacitance of capacitors C2 and C3(connected in series) be C'.

C’ = 100pF

Let the equivalent capacitance of capacitors C1 and C’(connected in parallel) be C’’.

C’’ = C’ + C1 = 100 + 100 = 200pF

Let the equivalent capacitance of C’’ and C4 (connected in series) be C’’’.

C’’’ = 200/3pF

Thus, the equivalent capacitance of the circuit = 200/3 pF.

Now, Supply potential (V) = 300 V

Charge on C4 is given by,

Q4 = C’’’V = = 2 × 10-8C

So, V4 = Q4/C4 = (210-8)/(10010-12) = 200C

Potential difference across C1 is given by, V1 = V-V4 = 300-200 = 100V

Charge on C1 is given by, Q1 = C1V1 = 100 × 10-12 × 100 = 10-8C

C2 and C3 having same capacitances have a potential difference of 100 V together.

Since C1 and C3 are in series, the potential difference across C2 and C3 is given by,V2 = V3 = 50V

Therefore charge on C2 is given by,Q2 = C2V2 = 20010-1250 = 10-8C

And charge on C3 is given by,Q3 = C3V3 = 20010-1250 = 10-8C.

The charge and voltage across each capacitor are as follows:

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