2.29 A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.34). Show that the capacitance of a spherical capacitor is given by
C =
where and are the radii of outer and inner spheres, respectively.
2.29 Let F be the force applied to separate the plates of a parallel plate capacitor and let be the distance.
Hence work done by the force = F
The potential energy increase in the capacitor = uA , where u = Energy density, A = area of each plate.
If d = distance between the plates and V = potential difference across the plates, the
Work done = increase in the potential energy.
Therefore,
F = uA or F = uA = ( )A
Electric intensity is given by
E =
F = ( )EA= ( )EA
Since Capacitance, C =
F = ( E) = QE
<p><strong>2.29 </strong>Let F be the force applied to separate the plates of a parallel plate capacitor and let <span title="Click to copy mathml"><math><mi>Δ</mi><mi>x</mi><mi></mi></math></span> be the distance.</p><p>Hence work done by the force = F <span title="Click to copy mathml"><math><mi></mi><mi>Δ</mi><mi>x</mi></math></span></p><p>The potential energy increase in the capacitor = uA <span title="Click to copy mathml"><math><mi></mi><mi>Δ</mi><mi>x</mi></math></span> , where u = Energy density, A = area of each plate.</p><p>If d = distance between the plates and V = potential difference across the plates, the</p><p>Work done = increase in the potential energy.</p><p>Therefore,</p><p>F <span title="Click to copy mathml"><math><mi></mi><mi>Δ</mi><mi>x</mi></math></span> = uA <span title="Click to copy mathml"><math><mi></mi><mi>Δ</mi><mi>x</mi></math></span> or F = uA = ( <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><msup><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span> )A</p><p>Electric intensity is given by</p><p>E = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>d</mi></mrow></mrow></mfrac></math></span></p><p>F = ( <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>E</mi></math></span> )EA= ( <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mfrac><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>d</mi></mrow></mrow></mfrac></math></span> )EA</p><p>Since Capacitance, C = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>A</mi></mrow></mrow><mrow><mrow><mi>d</mi></mrow></mrow></mfrac></math></span></p><p>F = ( <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mi>C</mi><mi>V</mi></math></span> E) = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math></span> QE</p>
It is the farad (F). The name came from Michael Faraday. SI Unit of Capacitance means the ability of a system to store an electric charge. One farad is the capacitance of a device that needs one coulomb of charge to provide one volt potential difference across it. Mathematically, it is represented by - 1 Farad = 1 Coulomb/Volt.
In JEE Main Physics, electric potential and capacitance chapter has a weightage of 3% to 6%. You can expect around 1 or 2 questions from this chapter which carries 4 to 8 marks.
In NEET Physics exam, the chapter electric potential and capacitance carries a weightage of 2% to 5%, which means you can expect one question out of 45 questions.
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