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iVinArrow [24]
3 years ago
11

An isolated parallel-plate capacitor has a surface charge density. If the space between the plates is filled with a material of

dielectric constant k = 3, find the induced surface charge density ind on the dielectric?
Physics
1 answer:
anygoal [31]3 years ago
3 0

Answer:

Explanation:

2\sqrt{34}

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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
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Answer:

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Explanation:

The magnitude of the charge is expresses as Q = CV

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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

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C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

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Using the formula for Electric field intensity

E = V/d

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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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