Part1 A parallel-plate capacitor has a plate area of 63 cm2 and a plate separati

Question

Part1

A parallel-plate capacitor has a plate area of

63 cm2 and a plate separation of 5.76 mm. A

potential difference of 7.74 V is applied across

the plates with only air between the plates.

What is the capacitance before the dielecrict is inserted?

Answer in units of F

Part 2

The battery is then disconnected, and a piece

of glass (with a dielectric constant 3.62) is

inserted to completely fill the space between

the plates.

What is the capacitance after the dielectric

is inserted?

Answer in units of F

Part 3

What is the charge on the plates before the

dielectric is inserted?

Answer in units of C

Part 4

What is the charge on the plates after the

dielectric is inserted?

Answer in units of C

Part 5

What is the potential difference across the

plates before the dielectric is inserted?

Answer in units of V

Part 6

What is the potential difference across the

plates after the dielectric is inserted?

Answer in units of V

Explanation / Answer

Capacitance C = charge stored/voltage across capacitor

C = e0*k*A/d where 1<= k, e0 = 8.85e-12 F/m, A = area of capacitor plate, d = separation between plates.

Assuming vacuum between plates, k = 1.

C = 8.85x10^-12 (F/m)*4*10^-4 (m^2)/(2.8x10^-3 (m))

C = 1.26x10^-12 F

So V = C/Q = 1.26x10^-12/(400x10^-12) = 3.2 mV

b. V = E*d so E = V/d = 3.2x10^-3/(2.8x10^-3) = 1.14 V/m

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