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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Capacitors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Capacitors

A spherical capacitor is formed by two conducting spherical shells...

A spherical capacitor is formed by two conducting spherical shells having radii aa and bb. The region between these shells contains a dielectric of dielectric constant KK up to a radius cc, as depicted in the figure. Determine the capacitance.
A spherical capacitor is formed by two conducting spherical shells having radii a and b . The region between these

Options

  1. A4πε0Kabca(b−c)+Kb(c−a)\dfrac{4\pi\varepsilon_0 K a b c}{a (b - c) + K b(c - a)}
  2. B4πε0KabcKa(b−c)+b(c−a)\dfrac{4\pi\varepsilon_0 K a b c}{K a (b - c) + b(c - a)}
  3. C4πε0KabcKb(c−a)+a(b−c)\dfrac{4\pi\varepsilon_0 K a b c}{K b (c - a) + a(b - c)}
  4. D4πε0abcKa(b−c)+b(c−a)\dfrac{4\pi\varepsilon_0 a b c}{K a (b - c) + b(c - a)}

Correct answer

B. 4πε0KabcKa(b−c)+b(c−a)\dfrac{4\pi\varepsilon_0 K a b c}{K a (b - c) + b(c - a)}

Step-by-step solution

The given spherical capacitor can be considered as a combination of two spherical capacitors connected in series:

1. A capacitor C1C_1 with a dielectric of constant KK between the radii aanda andc.<br>.<br>C_1 = \dfrac{4\pi\varepsilon_0 K a c}{c - a}<br><br>2.Acapacitor<br><br>2. A capacitorC_2withairbetweentheradiiwith air between the radiicandandb.<br>.<br>C_2 = \dfrac{4\pi\varepsilon_0 b c}{b - c}<br><br>Sincetheelectricfieldlinespassradiallythroughbothregions,thetwocapacitorsareinseries.Theequivalentcapacitance<br><br>Since the electric field lines pass radially through both regions, the two capacitors are in series. The equivalent capacitanceCisgivenby:<br><br>is given by:<br><br>\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}<br><br><br><br>\dfrac{1}{C} = \dfrac{c - a}{4\pi\varepsilon_0 K a c} + \dfrac{b - c}{4\pi\varepsilon_0 b c}<br><br><br><br>\dfrac{1}{C} = \dfrac{1}{4\pi\varepsilon_0 c} \left( \dfrac{c - a}{K a} + \dfrac{b - c}{b} \right)<br><br>\dfrac{1}{C} = \dfrac{1}{4\pi\varepsilon_0 c} \left( \dfrac{b(c - a) + K a(b - c)}{K a b} \right)

C=4πε0KabcKa(b−c)+b(c−a)C = \dfrac{4\pi\varepsilon_0 K a b c}{K a(b - c) + b(c - a)}
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