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Cloud [144]
3 years ago
10

If two similar large plates each of area having surface charge density is +a and -b are separated by a distance d in air find th

e expression for the potential difference and capacitance between them
Physics
1 answer:
leva [86]3 years ago
5 0

Answer:

a. V = (a - b)d/2ε₀ b. 2ε₀A/d

Explanation:

a. The potential difference between the plates

Using Gauss' law, we first find the electric field between the plates

ε₀∫E.dA = Q where Q = charge enclosed, E = electric field

Now Q = [a +(-b)]A = (a - b)A where + a and -b are the surface charge densities of the plates and A is the area of the plates.

ε₀∫E.dA = Q

ε₀∫EdAcos0 + ε₀∫EdAcos0 = (a - b)A

ε₀E∫dA + ε₀E∫dA = (a -b)A

ε₀EA + ε₀EA = (a -b)A

2ε₀EA = (a -b)A

E = (a - b)/2ε₀

We now find the potential difference, V between the plates from

V = ∫E.dl

V =E∫dl

V = Ed  where ∫dl = d the distance between the plates.

V = (a - b)d/2ε₀

b. The capacitance between them

Capacitance C = Q/V

= (a - b)A ÷ (a - b)d/2ε₀

= 2ε₀A/d

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Explanation:

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Two charges +q and -q are situated at certain distance, at the point exactly midway betwee them what happens on electric field a
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Explanation:

In science, charge, also recognized as electric charge, electrical charge, or electrostatic charge and expressed q, is a component of a unit of body that reveals the extent to which it has more or fewer electrons than protons.

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Among all the options, the option (c) Electric field is not zero but potential is zero is the correct answer.

Therefore, the option (b) Electric field is zero but potential is not zero is an incorrect solution

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Hence, the option (d) Electric field is not zero but potential is zero is an incorrect solution.

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