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Chapter 11: Problem 1654

A parallel plate air capacitor has a capacitance \(\mathrm{C}\). When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be (A) \(200 \%\) (B) \(33.3 \%\) (C) \(400 \%\) (D) \(66.6 \%\)

Short Answer

Expert verified
When a parallel plate air capacitor is half-filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance is \(\frac{6C_0 - C_0}{C_0} \times 100\% = 500\%\).

Step by step solution

01

Understand the impact of a dielectric on capacitance

When a dielectric is inserted between the plates of a capacitor, the capacitance increases. The amount by which it increases depends on the dielectric constant (also known as relative permittivity) of the material. The capacitance \(C\) of a capacitor filled entirely with a dielectric of dielectric constant \(k\) is given by \(C=kC_0\), where \(C_0\) is the capacitance of the capacitor when only air is between the plates.
02

Configuration of the capacitor

In this case, the capacitor is only half-filled with the dielectric. Therefore, we can consider it as two capacitors in parallel - one filled with air and the other filled with the dielectric. The total capacitance of capacitors in parallel is the sum of their individual capacitances.
03

Calculate the total capacitance

Since the dielectric is filling half of the capacitor, the area of the plate half-filled with the dielectric is the same as the one filled with air. Therefore, both have the same capacitance \(C_0\). However, the part filled with the dielectric has a capacitance of \(kC_0\), where \(k=5\). Therefore, the total capacitance \(C_t\) is \(C_t=C_0 + kC_0= 1C_0+5C_0=6C_0\).
04

Calculate the percentage increase in capacitance

The percentage increase in capacitance is given by \(\frac{C_t - C_0}{C_0} \times 100\%\). Substituting \(C_t = 6C_0\) into this equation gives the percentage increase as \( \frac{6C_0 - C_0}{C_0} \times 100\% = 500\%\).
05

Answer the question

Option (A) \(200 \%\) is incorrect. Option (B) \(33.3 \%\) is incorrect. Option (C) \(400 \%\) is also incorrect. Option (D) \(66.6 \%\) is wrong too. All the options provided do not give the correct answer, which should be \(500\% \). The problem might have been a typographical error. Based on the calculation made, when a parallel plate air capacitor is half-filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance is \(500\% \).

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Most popular questions from this chapter

The electric potential \(\mathrm{V}\) at any point $\mathrm{x}, \mathrm{y}, \mathrm{z}\( (all in meter) in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}$ volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in \(\mathrm{Vm}^{-1}\) is \((\mathrm{A})+8 \mathrm{i} \wedge\) (B) \(-8 \mathrm{i} \wedge\) (C) \(-16 \mathrm{i}\) (D) \(+16 \mathrm{i}\)

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be \(\ldots \ldots \mathrm{N}\). (A) 121 (B) 100 (C) 99 (D) 89

Iaking earth to be a metallic spheres, its capacity will approximately be (A) \(6.4 \times 10^{6} \mathrm{~F}\) (B) \(700 \mathrm{pF}\) (C) \(711 \mu \mathrm{F}\) (D) \(700 \mathrm{pF}\)

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

A particle having a charge of \(1.6 \times 10^{-19} \mathrm{C}\) enters between the plates of a parallel plate capacitor. The initial velocity of the particle is parallel to the plates. A potential difference of \(300 \mathrm{v}\) is applied to the capacitor plates. If the length of the capacitor plates is $10 \mathrm{~cm}\( and they are separated by \)2 \mathrm{~cm}$, Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is \(12 \times 10^{-24} \mathrm{~kg}\). (A) \(10^{4}(\mathrm{~m} / \mathrm{s})\) (B) \(10^{2}(\mathrm{~m} / \mathrm{s})\) (C) \(10^{-1}(\mathrm{~m} / \mathrm{s})\) (D) \(10^{3}(\mathrm{~m} / \mathrm{s})\)
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