Question: Two capacitors of capacitance 5 microfarad and 1 microfarad are having the same potential 100 volts. Now these capacitors are so connected that their plates with opposite polarity are joined together. Find (a) common potential (b) charge on the capacitor after connection (c) charge flow in redistribution? Explain in details.

Solution:

Concept: Capacitance is the ability of a body to store an electric charge. It is defined as the ratio of the charge stored to the potential difference across the plates of the capacitor.

Given Data:

Capacitance of 1st capacitor, C₁ = 5 µF

Capacitance of 2nd capacitor, C₂ = 1 µF

Potential difference across the capacitors, V = 100 V

(a) Common Potential:

When the plates with opposite polarity are joined together, the potential difference across the capacitors will be the same. Therefore, the common potential will be the average of the potential of the two capacitors.

Common Potential = (Potential of capacitor 1 + Potential of capacitor 2) / 2

Common Potential = (100 + 100) / 2 = 100 V

Therefore, the common potential is 100 V.

(b) Charge on the Capacitor after Connection:

When the plates with opposite polarity are joined together, the charge on the capacitor will redistribute. The total charge on the system will remain the same but the distribution will change.

The total charge on the system is given by:

Q = CV

where, C is the capacitance and V is the potential difference.

Charge on capacitor 1, Q₁ = C₁V = 5 µF × 100 V = 500 µC

Charge on capacitor 2, Q₂ = C₂V = 1 µF × 100 V = 100 µC

Total charge on the system before connection, Q_before = Q₁ + Q₂ = 500 µC + 100 µC = 600 µC

Total capacitance after connection, C_total = C₁ + C₂ = 5 µF + 1 µF = 6 µF

Using the formula Q = CV, the charge on the capacitor after connection can be calculated as:

Q_after = C_totalV = 6 µF × 100 V = 600 µC

Therefore, the charge on the capacitor after connection is 600 µC.

(c) Charge Flow in Redistribution:

When the plates with opposite polarity are joined together, the charge on the capacitors redistributes. The charge flows from the capacitor with higher potential to the capacitor with lower potential until the two capacitors have the same potential.

Charge on capacitor 1 before connection, Q_1,before = 500 µC

Charge on capacitor 2 before connection, Q_2,before

Q1= (C1)V = 5 micron×100volt = 500 micron C
Q2 = (C2)V = 1 micron× 100 volt = 100 micronC
total charge, Q = |Q1-Q2| = 400 micron C
As they are having same potential,so it is obvious that they are connected in parallel combination. Hence,
equivalent capacitance, C = C1 + C2 = 6 micron C.

(a).

Common Potential, V = total charge (Q)/equivalent capacitance (C) = 400 micronC/6 micronF = (200/3)V