Question

For a fixed plate separation and voltage, what happens to the stored energy as the plates are brought closer together?

		Increases
		It stays the same
		Decreases

For a fixed charge what happens to the capacitance of a parallel plate capacitor if the distance between the plates is increased?

		Increases
		It stays the same
		Decreases

For a fixed charge, what would happen to the capacitance of a parallel plate capacitor if the area of the plates were increased?

		It stays the same
		Increases
		Decreases

What happens to the capacitance if a conductor is placed between the plates?

		Decreases
		It stays the same
		Increases

What is the capacitance of a cylindrical coaxial capacitor with inner radius a=2cm, outer radius b= 4cm, and length L=10cm?

		2.5 pF
		-8 pF
		8 pF
		4 pF

What happens to the capacitance if a dielectric is inserted between the plates?

		Decreases
		Increases
		It doesn't change

A dielectric with dielectric constant 4 is inserted between a parallel plate capacitor filling it completely. If the capacitance before the inserting the dielectric was 100pF what is the new capacitance?

		400 pF
		100 pF
		25 pF
		1600 pF

For a fixed separation, what happens to the capacitance of a parallel plate capacitor if the charge is increased?

		It remains the same
		Increases
		Decreases

What is true about capacitance?

		It depends on both geometry and charge
		It only depends on geometry
		It only depends on the charge

What is the energy stored in a 1000nF capacitor if there are 2V applied?

		4x10-6 J
		2x10-6 J
		1.25x10-7 J
		1x10-6 J

Answer 1

1. Increases

Because by bringing the plates closer we are increasing the capacity of the capacitor hence the energy will also be increasing.

2. Decreases

Because the capacitance value of capacitor is inversely proportional to the distance between the plates as the distance increases the capacitor value decreases.

3. Increases

As the capacitance is directly proportional to the area of the plates so that as we increase the area our capacitance will also be increased.

4. Decreases

Because as the capacitor are now in series whose resultant gives us lesser value then the previous.

5. 8 pF

Capacitance will be given as,

C = 2 pie e × L/ln(b/a)

= 2×3.14×8.85×10^-12 × 0.1 /ln(4/2)

= 8 pF

6. Increase

As the relationship between the capacity of the capacitor and the dielectric strength is directly proportional so that when we insert di electric then our capacitance will increase accordingly.

7. 400 pF

Initially C = 100 pF

As the capacitor strengthens is directly proportional to the dielectric so that,

C' = 4 × C = 400 pF

8. Remains the same

As the capacitor strength is solely depends on the configuration or geometry of the capacitor nothing else.

9. It depends on geometry.

C = eA/d

So no charge depending here.

10. 2 × 10^-6 J

E = 0.5 C V^2 = 0.5 × 1000×10^-9 × 4 = 2×10^-6 J