Q. 795.0( 2 Votes )

# Find the charge on each of the capacitors 0.20 ms after the switch S is closed in figure.

Answer :

Concepts/Formulas used:

Charging a capacitor:

A capacitor of capacitance C is connected in series with a resistor of resistance R, a switch, and battery of emf ϵ . It is uncharged at first. The switch is closed at t = 0, then at time any time t the charge stored on the capacitor is given by

Capacitors in parallel:

If capacitors C_{1}, C_{2}, C_{3} , … are in parallel, then the equivalent capacitance is given by:

If the charges on the capacitors are Q1, Q2, Q3, .. are in parallel, then the charge on the capacitor with equivalent capacitance is given by:

We can replace the two capacitors by another capacitor of capacitance C. As the capacitors are in parallel.

Now,

We know that

Here,

Also, and

Hence,

Let the charge on both the capacitors be Q. As both have the same capacitance and potential (, both must have the same charge. Note that they both are in parallel.

Hence,

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A parallel-plate capacitor has plate area 20 cm^{2}, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy for the capacitor 8.9 μs after the connections are made.

How many time constants will elapse before the energy stored in the capacitor reaches half of its equilibrium value in a charging RC circuit?

HC Verma - Concepts of Physics Part 2