Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value Q1 such that the potential difference between the plates remains the same. 1) Compare Q1 and Q0.

Q1 < Q0

Q1 = Q0

Q1 > Q0

2) Compare the capacitance of the two configurations in the above problem.

C1 > C0

C1 = C0

C1 < C0

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Netta00 Netta00 · Answer 1 · 2019-11-01T07:59:13+01:00

Answer:

Q₁ > Qo

C₁ > Co

Explanation:

We know that

[tex]C_o=\dfrac{\varepsilon _oA}{d}[/tex]

Lets take relative permitivity of green plate =k

So new capacitance value

[tex]C_1=k\dfrac{\varepsilon _oA}{d}[/tex]

From above two expression we can say that

C₁ = kCo

k > 1

So we can say that

C₁ > Co

We also know that

Co = Qo V

C₁ = Q₁ V

From above we can say that

Co/C₁ = Qo/Q₁ ( C₁ > Co ⇒Co/C₁ <1)

Co/C₁ <1

So we can say that

Qo/Q₁ < 1

Q₁ > Qo