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dem82 [27]
2 years ago
11

A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6

V/m between the plates, the magnitude of the charge on each plate should be:
Physics
1 answer:
Oduvanchick [21]2 years ago
5 0

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

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Explanation:

First, consider the energy balance:

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The product m_{c} *Cp is the heat capacity, so the balance equation is:

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Answer:

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Explanation:

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λ = the linear charge density of the wire.

Now looking at the equatiom, due to the fact that Er varies inveresely with r, its maximum value will occur at the surface of the wire where r = R, the radius of the wire:

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