Question

1) A parallel plate capacitor is connected to a battery. The electric field between the plates is E. While still connected to the battery, we move the plates so that their plate separation is now twice as large. What is the electric field between the plates now?

E/4.

E.

E/2.

4E.

2E.

2)

A parallel plate capacitor with capacitance C_o is fully charged. The plates are in the shape of a disk. If the diameter of the disk is doubled and the separation between the plates is also doubled, what will the new capacitance be?

Group of answer choices

0.5 C_o

2 C_o

0.25 C_o

C_o

4 C_o

3)

Capacitors C₁ and C₂ are connected to battery in series with each other. While everything is connected, a dielectric is inserted into C₁. How does the charge change on C₂?

Group of answer choices

The charge on C₂ remains the same.

The charge on C₂ decreases.

The charge on C₂ increases.

Impossible to determine without knowing the dielectric constant.

Answer 1

Ans

1

As the battery is connected to capacitor plates so the potential difference across capacitor plates remains constant.

Let initially separation between plates is d thus electric field

Where V is the potential difference across the capacitor.

Now separation between plates is d' = 2d and the potential difference between plates remain the same because the battery is connected to the capacitor.

So electric field

2

The capacitance of the parallel plate capacitor is given by

Where A = Area of plates, d = distance between plates, = Permittivity of free space = 8.85 10^-12 F.m^-1

For disk

Where D = Diameter of the disk.

Thus

Now the Diameter of the disk is half. So D' = D/2 and separation between plates is also doubled so d' = 2d. Thus new capacitance is

C = 2 C_o

3

As Capacitor C₁ and C₂ are in series thus charge on both capacitors is equal to total charge on the equivalent capacitor.

Now dielectric is inserted between capacitor C₁ thus the capacitance of the capacitor C₁ increases and equivalent capacitance of capacitors C₁ and C₂ also increases. As the battery is connected so the potential difference across the equivalent capacitor remains the same.

As Q = C V

Where Q = Charge on the capacitor, V = Potential difference across the capacitor, C = Capacitance.

Now C increases but V remain the same so charge also increases

Therefore option The charge on C₂ increases. is correct.

Cheers