Physics, 14.07.2020 01:01, sterlingrobinson35

The plates of a parallel-plate capacitor are 3.22 mm apart, and each has an area of 9.72 cm^2. Each plate carries a charge of magnitude 4.45 times 10^-8 C. The plates are in vacuum. 1. What is the capacitance? Express your answer with the appropriate units.
2. What is the potential difference between the plates? Express your answer with the appropriate units.
3. What is the magnitude of the electric field between the plates? Express your answer with the appropriate units.

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answered: tajanaewilliams77

1. C = 2.67pF

2. V = 16.6kV

3. E = 5.17*10^6 N/C

Explanation:

1. The capacitance of the given parallel plate capacitor is calculated with the following formula:

$C=\frac{\epsilon_o A}{d}$ (1)

ε0: dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2

A: area of the plates = 9.72cm^2

d: distance between the plates = 3.22m = 3.22*10^-3m

You convert the area to m^2:

$A=9.72cm^2*\frac{(10^{-2}m)^2}{1cm^2}=9.72*10^{-4}m^2$

Then, replace the values of all parameters in the equation (1):

$C=\frac{(8.85*10^{-12}C^2/Nm^2)(9.72*10^{-4}m^2)}{3.22*10^{-3}m}=2.67*10^{-12}F\\\\C=2.67pF$

The capacitance of the parallel plate capacitor is 2.67pF

2. The potential difference between the plates is:

$V=\frac{Q}{C}$ (2)

Q: charge of each capacitor = 4.45*10^-8C

$V=\frac{4.45*10^{-8}C}{2.67*10^{-12}F}=1.66*10^4V\\\\V=16.6kV$

The potential difference between the plates is 16.6 kV

3. The magnitude of the electric field between the plates is:

$E=\frac{V}{d}$

$E=\frac{1.6*10^4V}{3.22*10^{-3}m}=5.17*10^6\frac{N}{C}$

The electric field between the plates has a magntiude of 5.17*10^6 N/C

answered: Guest

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answered: Guest

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