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rusak2 [61]
3 years ago
8

Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measure

d 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
Physics
1 answer:
jonny [76]3 years ago
3 0

Answer:

Q₁ > Qo

C₁ > Co

Explanation:

We know that

C_o=\dfrac{\varepsilon _oA}{d}

Lets take relative permitivity of green plate =k

So new capacitance value

C_1=k\dfrac{\varepsilon _oA}{d}

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

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Answer:

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Explanation:

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<u></u>

<u>FOR NUMBER OF P ATOMS</u>:

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<u>No. of Atoms of P = 3 atoms</u>

<u></u>

<u>FOR NUMBER OF Na ATOMS</u>:

No.\ of\ Atoms\ of\ Na = (No.\ of\ Moles)(N.\ of atoms of Na)\\No.\ of\ Atoms\ of\ Na = (3)(3\ atom})

<u>No. of Atoms of Na = 9 atoms</u>

<u></u>

<u>FOR TOTAL NUMBER OF ATOMS</u>:

Total\ No.\ of\ Atoms = (No.\ of\ Moles)(N.\ of\ atoms\ of\ Na + No.\ of\ atoms\ of\ P + No.\ of\ atoms\ of\ O)\\ Total\ No.\ of\ Atoms\ = (3)(3\ atoms + 1\ atom + 4\ atoms}) = (3)(8\ atoms)

<u>Total No. of Atoms = 24 atoms</u>

<u></u>

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The capacitance of two square parallel plates will be 1.777 farads.

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The body's capacity to hold electric charge is constrained by capacitance. There is a capacitance for each capacitor. Two metallic sheets of area A, spaced by d, make up the standard parallel-plate capacitor.

The parallel plate capacitor formula is given by:

C = ε Ad

Two square parallel plates 6.7 cm on a side are separated by 1.8 mm of paraffin. The dielectric constant of paraffin is 2.2. Then the capacitance of two square parallel plates will be given as,

C = ε Ad

C = 2.2 x 0.67 x 0.67 x 1.8

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Two parallel square plates will have a capacitance of 1.777 farads.

More about the capacitance of two parallel plates link is given below.

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