
The capacitance of a spherical condenser is \[1\mu F\] . If the spacing between the two spheres is \[\text{1 mm}\] , then the radius of the outer sphere is
(A) \[30cm\]
(B) \[6m\]
(C) \[5cm\]
(D) \[3m\]
Answer
227.1k+ views
Hint: A spherical capacitor consists of a solid or hollow spherical conductor of a certain radius surrounded by another concentric spherical conductor of a larger radius. The capacitance for spherical conductors can be obtained by finding the voltage difference between the conductors for a given charge.
Formula Used:
\[C=\dfrac{4\pi {{\varepsilon }_{0}}{{R}_{1}}{{R}_{2}}}{{{R}_{2}}-{{R}_{1}}}\]
Complete step by step answer:
We have been provided with the capacitance of the spherical capacitor and the spacing between the two conductors, which is the difference between their radii.
Capacitance of the spherical capacitor \[(C)=1\mu F={{10}^{-3}}F\] since \[1\mu F={{10}^{-6}}F\]
Also, spacing between the conductors \[({{R}_{2}}-{{R}_{1}})=1mm={{10}^{-3}}m\]
Now since the spacing is very small, we can consider the two capacitors to have almost equal radii, that is \[{{R}_{1}}\approx {{R}_{2}}\]
Also, we know the value of the constant in the capacitor formula, that is, \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\]
Substituting all the values listed above in our formula, we get
\[\begin{align}
& C=\dfrac{4\pi {{\varepsilon }_{0}}{{R}_{1}}{{R}_{2}}}{{{R}_{2}}-{{R}_{1}}} \\
& \Rightarrow {{10}^{-6}}F=\dfrac{{{R}_{1}}{{R}_{2}}}{9\times {{10}^{9}}\times {{10}^{-3}}} \\
& \Rightarrow {{R}_{1}}{{R}_{2}}=9{{m}^{2}} \\
& \Rightarrow {{R}_{2}}^{2}=9{{m}^{2}}(\because {{R}_{1}}\approx {{R}_{2}}) \\
& \Rightarrow {{R}_{2}}=3m \\
\end{align}\]
Hence the outer and the inner radius of the spherical conductors forming the capacitor are approximately equal to \[3m\] (a difference of one millimetre in their radius)
But we are only concerned with the outer radius and hence option (D) is the correct answer.
Additional Information: Every spherical conductor having a certain amount of charge acts as a capacitor, even an isolated sphere is considered as a capacitor whose second plate is at infinity. The applications for an isolated spherical capacitor or a pair of spherical capacitors illustrate that a charged sphere has some stored energy as a result of being charged.
Note: Although we have assumed the radii of the two spheres to be equal, we didn’t take their difference to be zero because the spacing between them, although very negligible, is still a gap and means that their radii are only approximately equal. We had to make the assumption because we didn’t have any other piece of information to help us solve the question.
Formula Used:
\[C=\dfrac{4\pi {{\varepsilon }_{0}}{{R}_{1}}{{R}_{2}}}{{{R}_{2}}-{{R}_{1}}}\]
Complete step by step answer:
We have been provided with the capacitance of the spherical capacitor and the spacing between the two conductors, which is the difference between their radii.
Capacitance of the spherical capacitor \[(C)=1\mu F={{10}^{-3}}F\] since \[1\mu F={{10}^{-6}}F\]
Also, spacing between the conductors \[({{R}_{2}}-{{R}_{1}})=1mm={{10}^{-3}}m\]
Now since the spacing is very small, we can consider the two capacitors to have almost equal radii, that is \[{{R}_{1}}\approx {{R}_{2}}\]
Also, we know the value of the constant in the capacitor formula, that is, \[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\]
Substituting all the values listed above in our formula, we get
\[\begin{align}
& C=\dfrac{4\pi {{\varepsilon }_{0}}{{R}_{1}}{{R}_{2}}}{{{R}_{2}}-{{R}_{1}}} \\
& \Rightarrow {{10}^{-6}}F=\dfrac{{{R}_{1}}{{R}_{2}}}{9\times {{10}^{9}}\times {{10}^{-3}}} \\
& \Rightarrow {{R}_{1}}{{R}_{2}}=9{{m}^{2}} \\
& \Rightarrow {{R}_{2}}^{2}=9{{m}^{2}}(\because {{R}_{1}}\approx {{R}_{2}}) \\
& \Rightarrow {{R}_{2}}=3m \\
\end{align}\]
Hence the outer and the inner radius of the spherical conductors forming the capacitor are approximately equal to \[3m\] (a difference of one millimetre in their radius)
But we are only concerned with the outer radius and hence option (D) is the correct answer.
Additional Information: Every spherical conductor having a certain amount of charge acts as a capacitor, even an isolated sphere is considered as a capacitor whose second plate is at infinity. The applications for an isolated spherical capacitor or a pair of spherical capacitors illustrate that a charged sphere has some stored energy as a result of being charged.
Note: Although we have assumed the radii of the two spheres to be equal, we didn’t take their difference to be zero because the spacing between them, although very negligible, is still a gap and means that their radii are only approximately equal. We had to make the assumption because we didn’t have any other piece of information to help us solve the question.
Recently Updated Pages
JEE Main Mock Test 2025-26: Electromagnetic Induction & Alternating Currents

JEE Main Mock Test 2025-26: Dual Nature of Matter & Radiation

JEE Main Mock Test 2025-26: Oscillations and Waves Free Practice

JEE Main Mock Test 2025-26: Thermodynamics Practice & Tips

JEE Main Mock Test 2025-26: Optics Chapter Practice Online

JEE Main 2025-26 Mock Test: Current Electricity Chapter Practice

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Derivation of Equation of Trajectory Explained for Students

Understanding Average and RMS Value in Electrical Circuits

Degree of Dissociation: Meaning, Formula, Calculation & Uses

