Problem 46 - Master Resource Book in JEE Main Physics

Question

Two identical spheres carrying charges \(-9 \mu \mathrm{C}\) and \(5 \mu \mathrm{C}\), respectively are kept in contract and then separated from each other. Point out true statement from the following in each sphere. (a) \(1.25 \times 10^{13}\) electrons are in excess (b) \(1.25 \times 10^{13}\) electrons are in deficit (c) \(4.15 \times 10^{12}\) electrons are in excess (d) None of the above

Step-by-Step Solution

Answer

(a) \(1.25 \times 10^{13}\) electrons are in excess.

Step 1: Find the Total Charge

The charges on the two identical spheres are \(-9 \mu \mathrm{C}\) and \(5 \mu \mathrm{C}\). When they come into contact, charges redistribute equally. Total charge is \(-9 \mu \mathrm{C} + 5 \mu \mathrm{C} = -4 \mu \mathrm{C}\).

Step 2: Redistribute the Charges

Since the spheres are identical, the charge will distribute equally. The charge on each sphere after separation becomes \(-\frac{4 \mu \mathrm{C}}{2} = -2 \mu \mathrm{C}\).

Step 3: Calculate Excess/Deficit of Electrons

To find how many electrons correspond to \(-2 \mu \mathrm{C}\), use the formula: \[ n = \frac{Q}{e} \]where \(Q = -2 \times 10^{-6} \mathrm{C}\) is the charge of one sphere, and \(e = 1.6 \times 10^{-19} \mathrm{C}\) is the elementary charge.\[ n = \frac{-2 \times 10^{-6}}{1.6 \times 10^{-19}} = -1.25 \times 10^{13} \text{ electrons} \]This means \(1.25 \times 10^{13}\) electrons are in excess because the charge is negative.

Key Concepts

Charge Redistribution

When two charged objects come into contact, their charges redistribute to achieve equilibrium. This means the total charge is shared equally between the objects.
In our exercise, two identical spheres are involved: one with a charge of -9 microcoulombs (\(-9 \mu C\)), and the other with a charge of 5 microcoulombs (\(5 \mu C\)).
When they touch, the total charge on both is calculated by simply adding their charges:

• Total Charge = \(-9 \mu C + 5 \mu C = -4 \mu C\)

Given the spheres are identical, this total charge of \(-4 \mu C\) distributes equally. Each sphere ends up with half of the total charge:

• Charge on each sphere = \(\frac{-4 \mu C}{2} = -2 \mu C\)

Charge redistribution is a fundamental concept, illustrating how charges move to balance themselves across connected objects.

Elementary Charge

The elementary charge is the smallest stable unit of electric charge and is denoted by \(e\). It has a value of \(1.6 \times 10^{-19}\) coulombs.
This constant is crucial when converting between measurable electric charge and the number of electrons involved. Since electrons carry a negative charge, their effect is usually discussed in terms of excess or deficit of the elementary charge.
Knowing the elementary charge allows us to calculate the number of electrons for a given charge on an object.
In this case:

• Each sphere has a charge of \(-2 \mu C\)
• We use the formula \( n = \frac{Q}{e} \) to find the equivalent number of electrons \(n\).
• \( Q = -2 \times 10^{-6}\) C
• Therefore, \(n = \frac{-2 \times 10^{-6}}{1.6 \times 10^{-19}}\)

This calculation shows us the significance of the elementary charge in quantifying an object's charge at the atomic level.

Excess and Deficit of Electrons

Electrons, as negatively charged particles, add or subtract from objects to adjust their net charge.
If an object has excess electrons, it is negatively charged, whereas a deficit means it has fewer electrons than positive charges and is positively charged.
For the identical spheres, after redistribution:

• Each has a charge of \(-2 \mu C\)
• This negative charge indicates an excess of electrons.
• Using the calculation: \(n = \frac{-2 \times 10^{-6}}{1.6 \times 10^{-19}}\)

The result is \(-1.25 \times 10^{13}\) electrons, showing that \(1.25 \times 10^{13}\) extra electrons are present.
This example reinforces how electric charge and the concept of electron transfer explains excess or deficit conditions, adjusting the net charge on an object correspondingly.