
Considering earth as a metallic sphere, its capacitance would be nearly __________$\mu F.$ (Radius of earth $ = 6400{\text{ km, }}{ \in _0} = 8.85 \times {10^{ - 12}}$ SI unit)
$
{\text{A) 71}} \\
{\text{B) 7}}{\text{.11}} \\
{\text{C) 711}} \\
{\text{D) 7}}{\text{.1}} \times {\text{1}}{{\text{0}}^5} \\
$
Answer
580.8k+ views
Hint: Capacitance is the ability of the capacitor to collect and store energy in the form of the electrical charge. Capacitors are an electronic component which stores energy with the charge and potential difference across the terminals which are available in many sizes and shapes. To find the capacitance of the metallic sphere, use formula$c = 4\pi { \in _0}R$. Place the given values and find the unknown term.
Complete step by step answer:
Given that, Radius of the earth, $R = 6400{\text{ km}}$
Convert the given unit of radius in MKS system (Meter Kilogram Second) system.
Since, one kilometre is equal to thousand meters.
$
R = 6400km \\
\implies R = 6400 \times 1000m \\
$
${ \in _0} = 8.85 \times {10^{ - 12}}{\text{ SI unit}}$
According to the formula, the capacitance for the metallic sphere is $c = 4\pi { \in _0}R$
Place the given values in the above equations.
$c = 4 \times 3.14 \times 8.85 \times {10^{ - 12}} \times 6400 \times 1000$
Simplify the left hand side of the equation and find the value for the capacitance.
\[c = 4 \times \dfrac{{3.14}}{{{{10}^{12}}}} \times 8.85 \times 6400 \times 1000\] (By the law of exponents, when the exponents with negative powers are moved to denominator from the numerator becomes positive and vice-versa)
\[c = \dfrac{{7113.984 \times {{10}^5}}}{{{{10}^{12}}}}\]
Use, the law of exponents – the exponents with the same base and in division, then powers are subtracted)
$
c = \dfrac{{7113.984}}{{{{10}^{12 - 5}}}} \\
\implies c = \dfrac{{7113.984}}{{{{10}^7}}} \\
$
Now, as per the required answer is in micro farad, convert the above expression in that form –
$c = \dfrac{{7113.984}}{{{{10}^6} \times 10}}$
Take denominator to the numerator to make the given exponent in micro terms
$
c = \left( {\dfrac{{7113.984}}{{10}}} \right) \times {10^{ - 6}} \\
\implies c = 711.3 \times {10^{ - 6}} \\
\therefore c = 711\mu F \\
$
Therefore, the required answer is – the capacitance would be nearly \[711\mu F\]
So, the correct answer is “Option C”.
Note:
The capacitance value of the capacitor is measured in farads (F) behind the name of the Physicist Michael Faraday. Since farad is the large quantity of the capacitance, most of the house-hold appliances use a fraction of farads and generally from microfarads to pico-farad.
Complete step by step answer:
Given that, Radius of the earth, $R = 6400{\text{ km}}$
Convert the given unit of radius in MKS system (Meter Kilogram Second) system.
Since, one kilometre is equal to thousand meters.
$
R = 6400km \\
\implies R = 6400 \times 1000m \\
$
${ \in _0} = 8.85 \times {10^{ - 12}}{\text{ SI unit}}$
According to the formula, the capacitance for the metallic sphere is $c = 4\pi { \in _0}R$
Place the given values in the above equations.
$c = 4 \times 3.14 \times 8.85 \times {10^{ - 12}} \times 6400 \times 1000$
Simplify the left hand side of the equation and find the value for the capacitance.
\[c = 4 \times \dfrac{{3.14}}{{{{10}^{12}}}} \times 8.85 \times 6400 \times 1000\] (By the law of exponents, when the exponents with negative powers are moved to denominator from the numerator becomes positive and vice-versa)
\[c = \dfrac{{7113.984 \times {{10}^5}}}{{{{10}^{12}}}}\]
Use, the law of exponents – the exponents with the same base and in division, then powers are subtracted)
$
c = \dfrac{{7113.984}}{{{{10}^{12 - 5}}}} \\
\implies c = \dfrac{{7113.984}}{{{{10}^7}}} \\
$
Now, as per the required answer is in micro farad, convert the above expression in that form –
$c = \dfrac{{7113.984}}{{{{10}^6} \times 10}}$
Take denominator to the numerator to make the given exponent in micro terms
$
c = \left( {\dfrac{{7113.984}}{{10}}} \right) \times {10^{ - 6}} \\
\implies c = 711.3 \times {10^{ - 6}} \\
\therefore c = 711\mu F \\
$
Therefore, the required answer is – the capacitance would be nearly \[711\mu F\]
So, the correct answer is “Option C”.
Note:
The capacitance value of the capacitor is measured in farads (F) behind the name of the Physicist Michael Faraday. Since farad is the large quantity of the capacitance, most of the house-hold appliances use a fraction of farads and generally from microfarads to pico-farad.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Why cannot DNA pass through cell membranes class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

In a human foetus the limbs and digits develop after class 12 biology CBSE

AABbCc genotype forms how many types of gametes a 4 class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The correct structure of ethylenediaminetetraacetic class 12 chemistry CBSE

