Sigma Percentile
LEVELBoard

Animated Solution for Physics - Electrostatics: Capacitance (in F) of a spherical conductor having radius , is

Select Answer:

Visualized Solution

Visualizing the Spherical Conductor

  • Let's consider a spherical conductor of radius .

Formula for Capacitance

  • The capacitance of an isolated spherical conductor of radius is given by:

Substituting the Values

  • We know the value of the electrostatic constant:
  • Therefore,
  • Substitute :

Calculating the Value

Final Answer

  • Adjusting the decimal to match the options:

Food for Thought

  • What if the sphere was placed in a dielectric medium of constant ?
  • The new capacitance would be .

The Sigma Insight: Capacitance and Capacitors

Solution Diagram

The Concept of an Isolated Capacitor

When we hear the word "capacitor," we usually picture two parallel plates separated by some distance. But did you know that a single, isolated conductor can also act as a capacitor?
Imagine a perfectly spherical conductor floating in the vast emptiness of free space. Even though there isn't a second plate nearby, this sphere can still store electric charge. In physics, we consider the "second plate" of this isolated capacitor to be located at infinity, where the electric potential is zero.

The Master Equation

The capacitance of any object tells us how much charge it can store for a given rise in potential. For an isolated spherical conductor of radius , the capacitance is given by a beautifully simple formula:
Notice something fascinating here? The capacitance depends only on the geometry of the conductor (its radius ) and the medium surrounding it (represented by the permittivity ). It doesn't matter if the sphere is made of copper, gold, or aluminum; its ability to store charge remains exactly the same!

Executing the Calculation

In our problem, we are given a spherical conductor with a radius . Let's plug this into our formula.
We know the value of the electrostatic constant:
Therefore, the term is simply the reciprocal:
Now, substituting :
Let's compute this value. Dividing by gives approximately . Bringing the from the denominator to the numerator changes the sign of the exponent:

The Final Touch

To match our answer with the given options, we need to adjust the decimal point. By shifting the decimal one place to the right, we must decrease the exponent by one:
This perfectly matches option (a).
Food for thought: A capacitance of (or ) is incredibly small! This shows just how massive the unit of "1 Farad" truly is. To have a capacitance of 1 Farad, an isolated sphere would need a radius of —that's roughly 13 times the radius of our Sun!

Similar Questions

JEE Main 2019
LEVELJEE Main

A parallel plate capacitor with plates of area each, are at a separation of . If the electric field between the plates is , the magnitude of charge on each plate is

(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Advanced

A parallel plate capacitor is made of two square plates of side '' separated by a distance (). The lower triangular portions is filled with a dielectric of dielectric constant , as shown in the figure. Capacitance of this capacitor is

(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main

In a parallel plate capacitor set up, the plate area of capacitor is and the plates are separated by . If the space between the plates are filled with a dielectric material of thickness and area (see figure) the capacitance of the set-up will be ...... . (Dielectric constant of the material ) (Round off to the nearest integer)

JEE Main 2021
LEVELJEE Main

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Take, area of plate = )

(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Main

Voltage rating of a parallel plate capacitor is . Its dielectric can withstand a maximum electric field of . The plate area is . What is the dielectric constant, if the capacitance is ? (Take, )

(A)
3.8
(B)
8.5
(C)
4.5
(D)
6.2
JEE Main 2020
LEVELJEE Advanced

A capacitor is made of two square plates each of side making a very small angle between them, as shown in figure. The capacitance will be close to

(A)
(B)
(C)
(D)
JEE Main 2014
LEVELJEE Main

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is V/m, the charge density of the positive plate will be close to

(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main

A parallel plate capacitor has plate area and plate separation of 10 m. The space between the plates is filled upto a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is . The value of . The value of to the nearest integer is …… .

JEE Main 2021
LEVELJEE Main

If is the free charge on the capacitor plates and is the bound charge on the dielectric slab of dielectric constant placed between the capacitor plates, then bound charge can be expressed as

(A)
(B)
(C)
(D)
JEE Advanced 1996
LEVELJEE Advanced

The capacitance of a parallel plate capacitor with plate area and separation , is . The space between the plates is filled with two wedges of dielectric constants and respectively (figure). Find the capacitance of the resulting capacitor.