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11 June, 19:04

A parallel-plate capacitor is constructed of two square plates, size L*L, separated by distance d. The plates are given charge ±Q.

Part A

What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if Qis doubled?

Part BWhat is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if Lis doubled? Part CWhat is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if d is doubled?

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Answers (1)
  1. 11 June, 20:47
    0
    Answer: A) 2 B) 4 C) 1

    Explanation:

    The Electric field from a parallel-plate capacitor is given by:

    A) E=Q / (L^2 * ε0) so if we put a charge double the final electric field is double that the original.

    B) from the above expression for the electric field, If the size of the plate is double, then the E final is four times weaker that the original.

    C) If the distante between plates is doubled the final electric field is the same that initial.
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