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natulia [17]
3 years ago
8

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

±Q.
a. What is the ratio Ef/Ei of the final to initial electric field strengths if L is doubled?
Physics
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

Explanation:

Given

Area of capacitor Plates A=L\times L

distance between plates is d

capacitance C is given by

C=\frac{\epsilon A}{d}

C=\frac{\epsilon \cdot L^2}{d}

Provided V is Voltage

Charge(Q)=capacitance(C)\times Voltage(V)

If L is doubled

Capacitance C'=\frac{\epsilon \cdot (2L)^2}{d}

C'=4\times \frac{\epsilon \cdot L^2}{d}

Electric field is given by

E=\frac{Q}{\epsilon _0A}

E_i=\frac{Q}{\epsilon _0L^2}---1

E_f=\frac{Q}{\epsilon _0(2L)^2}---2

divide 1 and 2 we get

\frac{E_i}{E_f}=\frac{(2L)^2}{L^2}

\frac{E_f}{E_i}=\frac}{1}{4}

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Explanation:

given,

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From Gauss law

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\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}

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Also given that r_1= smaller radius

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