A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material that fills the volume between the plates has a dielectric constant of 5.00. The plates of the capacitor are connected to a 400 V battery.

What is the capacitance of the capacitor?

What is the charge on either plate?

How much energy is stored in the charged capacitor?

Explanation:

Area of plate

A = 17cm² = 0.00017m²

Separation distance

d = 0.15cm = 0.0015m

Dielectric material k = 5

The plate of the capacitor are connected to

Voltage =400V

a. Capacitance of the capacitor?

Capacitance of the capacitor can be determine by

C = kεA/d

ε = 8.854 × 10^-12F/m

C=5×8.854×10^-12×0.00017/0.0015

C = 5.017 × 10^-12 F

C = 5.017 pF.

The capacitance of the capacitor is 5.017 picoFarad

b. The charge on a capacitor is given as

q = CV

q = 5.017 ×10^-12 × 400

q = 2.007 × 10^-9

q = 2.007 nC

Each plate has a charge of 2.007nC but one is positive and the other is negative

Q1 = +2.007nC

Q2 = - 2.007nC

So the charge on each plate is ±2.007 nano Coulombs

c. Energy stored in a capacitor is given as

E = ½CV²

E = ½ × 5.017×10^-12 × 400²

E = 4.0136 ×10^-7 J

So the energy stored by the capacitor is 4.0136 × 10^-7 J.

t2 equal the displacement of the particle over the same time period if t2> t/2? hint: consider a sine wave starting at t1=0, then consider a closing wave starting at t1=0, for your explanatio