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6 March, 12:51

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 the expression for the potential difference and capacitance between them

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  1. 6 March, 15:05
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    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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