Question

Two capacitors are 2and3 uF and are joined in series. The outer plate of the first capacitor is at 1000V and the outer plate of the second capacitor is earthed. Find out the potential and charge of the inner plate of each capacitor

Answer 1

Answer:

● Answer -

Q1 = 1200 μC, V1 = 600 V

Q2 = 1200 μC, V2 = 400 V

◆ Explaination -

When two capacitors atr connected in series, equivalent capacitance is given by -

C = C1.C2 / (C1+C2)

C = 2 × 3 / (2 + 3)

C = 6 / 5

C = 1.2 μF

As capacitors are connected in series, charge will be same.

Q = Q1 = Q2

Q = C.V

Q = 1.2 × 1000

Q = 1200 μC

Potential of inner plate of 2 μF capacitor -

V1 = Q / C1

V1 = 1200 / 2

V1 = 600 V

Potential of inner plate of 3 μF capacitor -

V2 = Q / C2

V2 = 1200 / 3

V2 = 400 V

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Answer 2

here is my answer

Explanation:

When two capacitors of 2 and 3 microfarad are connected in series then the equivalent capacitance will be ….

C (equivalent) = { (2×10^-6)×(3×10^-6)} / {(2×10^-6)+(3×10^-6)}

C(equivalent) = ( 1.2 × 10^-6 )

Now we have to measure the voltage drops across each capacitor …

Voltage drop across the 2 microfarad capacitor will be = { V(source) × C(equivalent)/C1 }

Vc1 = {( 1000× 1.2× 10^-6 ) /( 2 × 10^-6)}

Vc1 = 600V.

Voltage drop across the 3 microfarad capacitor is..

Vc2 = { ( 1000 × 1.2 × 10^-6 )/( 3 × 10^-6) }

Vc2 = 400 V.

Let's assume the outer plate of 3 microfarad capacitor is connected to the earth then the voltage across it will be 400V and the charge will be = ( 3 × 10^-6 ) × 400 = 1200× 10^-6 = 1.2× 10^-3.

Hope it will help you..

Thanks for reading !!!