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Question: The capacitance of a spherical conductor with radius 1 m is : A. \(9 \times {10^9}F\) B. \(1\mu ...

The capacitance of a spherical conductor with radius 1 m is :
A. 9×109F9 \times {10^9}F
B. 1μf1\mu f
C. 1.1×10−10F1.1 \times {10^{ - 10}}F
D. 1×10−8F1 \times {10^{ - 8}}F

Explanation

Solution

Hint: Spherical capacitors are spherical in shape therefore capacitance of a spherical conductor is given by the formula,
C=4π∈0RC = 4\pi { \in _{_0}}R
Where ∈0={ \in _0} = permittivity of free space
R = Radius of conductor
C = Capacitance of conductor
Capacitance of a spherical conductor can easily be calculated by finding the potential of the spherical conductor.

Complete step-by-step answer:
Given: Radius of capacitor = 1mm
We know that,
Electric field due to a spherical capacitance is given by,
E=Q4π∈0R2E = \dfrac{Q}{{4\pi { \in _0}{R^2}}}
We know that,
V=E×R⇔V=Q4π∈0R2×RV = E \times R \Leftrightarrow V = \dfrac{Q}{{4\pi { \in _0}{R^2}}} \times R
∴V=Q4π∈0R⋅⋅⋅⋅⋅⋅(1)\therefore V = \dfrac{Q}{{4\pi { \in _0}R}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)
Also we know that,
C=QV⋅⋅⋅⋅⋅⋅(2)C = \dfrac{Q}{V} \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)
From (1) and (2) we get
C=QQ4π∈0R⇔C=4π∈0RC = \dfrac{Q}{{\dfrac{Q}{{4\pi { \in _0}R}}}} \Leftrightarrow C = 4\pi { \in _0}R
Hence spherical capacitance of a spherical conductor given by
C=4π∈0rC = 4\pi { \in _0}r
But 14π∈0=9×109\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}
∴4π∈0=9×10−9\therefore 4\pi { \in _0} = 9 \times {10^{ - 9}} ∴4π∈0=19×10−9\therefore 4\pi { \in _0} = \dfrac{1}{9} \times {10^{ - 9}}
∴\therefore Required capacitance of conductor
C=19×10−9×(1)C = \dfrac{1}{9} \times {10^{ - 9}} \times (1)
On solving we get
C=1.1×10−10FC = 1.1 \times {10^{ - 10}}F
Hence the spherical capacitance of a conductor of radius 1mmis 1.1×10−101.1 \times {10^{ - 10}}

Note: A spherical capacitor consist of a hollow or a solid spherical conductor surrounded by another concentric hollow spherical conductor is given by
C=4π∈0r1r2r1−r2C = \dfrac{{4\pi { \in _0}{r_1}{r_{{2_{}}}}}}{{{r_1} - {r_2}}}
where r1{r_1}= outer radius of conductor
r2{r_{_2}}= inner radius of conductor
Relative permeability is used for capacitor having dielectric constant k, given by
∈r=∈0k{ \in _r} = { \in _0}k.