
Two identical spheres carrying charges - 9 uC and 5 uC respectively are kept in contact and then separated from each other. Point out the true statement from the following. In each sphere
1) 1.25 $\times$ ${10^{13}}$ electrons are in deficit.
2) 1.25 $\times$ ${10^{13}}$ electrons are in excess.
3) 2.15 $\times$ ${10^{13}}$ electrons are in excess.
4) 2.15 $\times$ ${10^{13}}$ electrons are in deficit.
Answer
220.2k+ views
Hint When two identical spheres are brought in contact to each other, the charges redistribute. And final charge on each sphere is given by Q = \[\dfrac{{{{\text{q}}_{\text{1}}}{\text{ + }}{{\text{q}}_{\text{2}}}}}{2}\]. Hence, now find out the number of electrons by using the formula = Magnitude of charge on each sphere / Charge on each electron to obtain the answer.
Complete step-by-step answer:
Given,
Two identical spheres carrying charges ${{\text{q}}_{\text{1}}}$ = -9 μC and ${{\text{q}}_{\text{2}}}$ = 5 μC respectively are kept in contact and then separated from each other.
When two identical spheres are kept in contact and then separated, their charges redistribute.
To find whether the electrons are in excess or deficit, we have to find out the net charge on the system i.e. Final charge on each sphere after they are separated.
Final charge on each of sphere, Q = \[\dfrac{{{{\text{q}}_{\text{1}}}{\text{ + }}{{\text{q}}_{\text{2}}}}}{2} = \dfrac{{ - 9 + 5}}{2} = - 2\mu C\]
Negative signs indicate that charge on each sphere is in excess . (Why? See Note at the end of answer)
So, number of electrons are in excess is given by,
Magnitude of charge on each sphere / Charge on each electron. Hence, substituting values we have :
$
\dfrac{{2\mu C}}{{1.6 \times {{10}^{ - 19}}}} \\
\Rightarrow \dfrac{{2 \times {{10}^{ - 6}}C}}{{1.6 \times {{10}^{ - 19}}}} \\
\Rightarrow 1.25 \times {10^{ - 13}} \\
$
Therefore 1.25 $\times$ ${10^{13}}$ electrons are in excess.
Note Generally, we consider electrons excess when the overall charge (Q) on each sphere has a negative sign and since charge on electrons is negative; hence we say that electrons are in excess. Had they been positive, we could say that electrons are in deficit.
Complete step-by-step answer:
Given,
Two identical spheres carrying charges ${{\text{q}}_{\text{1}}}$ = -9 μC and ${{\text{q}}_{\text{2}}}$ = 5 μC respectively are kept in contact and then separated from each other.
When two identical spheres are kept in contact and then separated, their charges redistribute.
To find whether the electrons are in excess or deficit, we have to find out the net charge on the system i.e. Final charge on each sphere after they are separated.
Final charge on each of sphere, Q = \[\dfrac{{{{\text{q}}_{\text{1}}}{\text{ + }}{{\text{q}}_{\text{2}}}}}{2} = \dfrac{{ - 9 + 5}}{2} = - 2\mu C\]
Negative signs indicate that charge on each sphere is in excess . (Why? See Note at the end of answer)
So, number of electrons are in excess is given by,
Magnitude of charge on each sphere / Charge on each electron. Hence, substituting values we have :
$
\dfrac{{2\mu C}}{{1.6 \times {{10}^{ - 19}}}} \\
\Rightarrow \dfrac{{2 \times {{10}^{ - 6}}C}}{{1.6 \times {{10}^{ - 19}}}} \\
\Rightarrow 1.25 \times {10^{ - 13}} \\
$
Therefore 1.25 $\times$ ${10^{13}}$ electrons are in excess.
Note Generally, we consider electrons excess when the overall charge (Q) on each sphere has a negative sign and since charge on electrons is negative; hence we say that electrons are in excess. Had they been positive, we could say that electrons are in deficit.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

