Question

A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.

Answer 1

Answer:

I don't know the answer

Explanation:

18ohm is correct

Answer 2

The electric field in the capacitor 10 ns after the connections are made is 1.7$\times$104 V/m.

Explanation:

It is shown that

Plates area, A = 40 $\times$ 10−4 m2

Between the plates, the separation is d = 0.1 mm  

Resistance, R = 16 Ω

Batter’s Emf,

V0 = 2 V

The parallel plate capacitor’s capacitance C,

C = ∈0 Ad = 8.85$\times 40 \times 10-12 \times 10-41 \times 10-4 = 35.4 \times 10-11 F$

So, the electric field,

E = Vd = QCd = Q0A∈0 1-e-tRC = 1.655 $\times 104 = 1.7 \times 104$ V/m