A metallic sphere of radius $18\,cm $ has been given a charge of $5 \times {10^{ - 6}}\,C $. The energy of the charged conductor is:
(A) $0.2\,J $
(B) $0.6\,J $
(C) $1.2\,J $
(D) $2.4\,J $
Answer
273.6k+ views
Hint The energy of the charge can be determined by using the two formulas, the potential difference of the capacitor at a distance formula and then the energy of the charged conductor formula. By using the potential difference of the capacitor formula, the capacitance is determined and then the energy is determined by the energy formula.
Useful formula
The potential difference of the capacitor at a distance is given by,
$V = \dfrac{{kQ}}{r} $
Where, $V $ is the potential difference, $k $ is the constant, $Q $ is the charge of the conductor and $r $ is the radius.
The energy of the charged conductor is given by,
$U = \dfrac{{{q^2}}}{{2C}} $
Where, $U $ is the energy of the charged conductor, $q $ is the charge of the conductor and $C $ is the capacitance.
Complete step by step answer
Given that,
A metallic sphere of radius is, $18\,cm = 18 \times {10^{ - 2}}\,m $
The charge of the metallic sphere is, $Q = q = 5 \times {10^{ - 6}}\,C $
Now,
The potential difference of the capacitor at a distance is given by,
$V = \dfrac{{kQ}}{r} $
By rearranging the terms in the above equation, then the above equation is written as,
$\dfrac{Q}{V} = \dfrac{r}{k} $
Now the term $\dfrac{Q}{V} $ is equal to the capacitance of the metallic sphere, then the above equation is written as,
$C = \dfrac{r}{k} $
By substituting the radius value and the constant value of the $k = 9 \times {10^9} $ in the above equation, then the above equation is written as,
$C = \dfrac{{18 \times {{10}^{ - 2}}}}{{9 \times {{10}^9}}} $
By dividing the terms in the above equation, then the above equation is written as,
$C = 2 \times {10^{ - 11}} $
Now,
The energy of the charged conductor is given by,
$U = \dfrac{{{q^2}}}{{2C}} $
By substituting the charge and the capacitance in the above equation, then the above equation is written as,
$U = \dfrac{{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}}}{{2 \times 2 \times {{10}^{ - 11}}}} $
By squaring the terms in the above equation, then the above equation is written as,
$U = \dfrac{{2.5 \times {{10}^{ - 11}}}}{{2 \times 2 \times {{10}^{ - 11}}}} $
By cancelling the same terms in the above equation, then the above equation is written as,
$U = \dfrac{{2.5}}{{2 \times 2}} $
By multiplying the terms in the above equation, then the above equation is written as,
$U = \dfrac{{2.5}}{4} $
By dividing the terms in the above equation, then the above equation is written as,
$U = 0.625\,J $
Thus, the above equation shows the energy of the metallic sphere.
Hence, the option (B) is the correct answer.
Note The capacitance of the conductor is directly proportional to the charge of the conductor and inversely proportional to the potential difference of the conductor. As the charge of the conductor then the capacitance of the conductor also increases.
Useful formula
The potential difference of the capacitor at a distance is given by,
$V = \dfrac{{kQ}}{r} $
Where, $V $ is the potential difference, $k $ is the constant, $Q $ is the charge of the conductor and $r $ is the radius.
The energy of the charged conductor is given by,
$U = \dfrac{{{q^2}}}{{2C}} $
Where, $U $ is the energy of the charged conductor, $q $ is the charge of the conductor and $C $ is the capacitance.
Complete step by step answer
Given that,
A metallic sphere of radius is, $18\,cm = 18 \times {10^{ - 2}}\,m $
The charge of the metallic sphere is, $Q = q = 5 \times {10^{ - 6}}\,C $
Now,
The potential difference of the capacitor at a distance is given by,
$V = \dfrac{{kQ}}{r} $
By rearranging the terms in the above equation, then the above equation is written as,
$\dfrac{Q}{V} = \dfrac{r}{k} $
Now the term $\dfrac{Q}{V} $ is equal to the capacitance of the metallic sphere, then the above equation is written as,
$C = \dfrac{r}{k} $
By substituting the radius value and the constant value of the $k = 9 \times {10^9} $ in the above equation, then the above equation is written as,
$C = \dfrac{{18 \times {{10}^{ - 2}}}}{{9 \times {{10}^9}}} $
By dividing the terms in the above equation, then the above equation is written as,
$C = 2 \times {10^{ - 11}} $
Now,
The energy of the charged conductor is given by,
$U = \dfrac{{{q^2}}}{{2C}} $
By substituting the charge and the capacitance in the above equation, then the above equation is written as,
$U = \dfrac{{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}}}{{2 \times 2 \times {{10}^{ - 11}}}} $
By squaring the terms in the above equation, then the above equation is written as,
$U = \dfrac{{2.5 \times {{10}^{ - 11}}}}{{2 \times 2 \times {{10}^{ - 11}}}} $
By cancelling the same terms in the above equation, then the above equation is written as,
$U = \dfrac{{2.5}}{{2 \times 2}} $
By multiplying the terms in the above equation, then the above equation is written as,
$U = \dfrac{{2.5}}{4} $
By dividing the terms in the above equation, then the above equation is written as,
$U = 0.625\,J $
Thus, the above equation shows the energy of the metallic sphere.
Hence, the option (B) is the correct answer.
Note The capacitance of the conductor is directly proportional to the charge of the conductor and inversely proportional to the potential difference of the conductor. As the charge of the conductor then the capacitance of the conductor also increases.
Recently Updated Pages
Mass vs Weight: Key Differences Explained for Students

Circuit Switching vs Packet Switching: Key Differences Explained

Uniform Acceleration Explained: Formula, Examples & Graphs

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Area vs Volume: Key Differences Explained for Students

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

