Question

Two small conducting sphere of equal radius have charges \(+1 \mathrm{c}\) and \(-2 \mathrm{c}\) respectively and placed at a distance \(\mathrm{d}\) from each other experience force \(F_{1}\). If they are brought in contact and separated to the same distance, they experience force \(F_{2}\). The ratio of \(F_{1}\) to \(F_{2}\) is \(\ldots \ldots \ldots \ldots\) (A) \(-8: 1\) (B) \(1: 2\) (C) \(1: 8\) (D) \(-2: 1\)

Short answer

The ratio of \(F_1\) to \(F_2\) is \(-8:1\).

Step-by-step solution

Calculate force F1

Use Coulomb's Law formula for the force between two charged objects: \(F_1 = k \frac{Q_1 Q_2}{d^2}\) where, k is Coulomb's constant = \(9 \times 10^9 Nm^2C^{-2}\), \(Q_1 = +1c\), \(Q_2 = -2c\), and d is the distance between the spheres. Plug in the values: \(F_1 = 9 \times 10^9 \frac{(+1)(-2)}{d^2}\)

Determine redistributed charges

When the spheres come into contact, the total charge becomes, \(Q_{total} = Q_1 + Q_2 = +1c - 2c = -1c\) Since the spheres have equal radius, the charge will be distributed equally between them after touching, so each sphere gets half of the total charge: \(Q_{1(new)} = Q_{2(new)} = -\frac{1}{2}c\)

Calculate force F2

Use Coulomb's Law formula again, now with the redistributed charges of the spheres. \(F_2 = k \frac{Q_{1(new)} Q_{2(new)}}{d^2}\) Plug in the values: \(F_2 = 9 \times 10^9 \frac{(-\frac{1}{2})(-\frac{1}{2})}{d^2}\)

Calculate the ratio of F1 to F2

Now, we need the ratio of F1 to F2. We can cancel the common factors on both sides, so we get: \(\frac{F_1}{F_2} = \frac{[-2]}{[\frac{1}{4}]}\) Solve it: \(\frac{F_1}{F_2} = -8\)

Select the correct option

The ratio of F1 to F2 is -8:1. So, the correct answer is: (A) -8:1

Key concepts

Charge Distribution

Charge distribution refers to how electric charge is spread across different areas or objects. When dealing with conducting spheres, interesting things happen when they come into contact. Initially, we had two charged spheres with charges of \(+1 \, ext{C}\) and \(-2 \, ext{C}\). When these two spheres touch, their charges redistribute so that the total charge is shared equally between them.
This happens because the spheres are conductors, meaning they allow charges to move freely across their surface. When the spheres come into contact, they act as a single conductor, allowing the charges to spread evenly between them until they are separated again. The total charge of \-1 \, ext{C}\ results in each sphere holding \(-\frac{1}{2} \, ext{C}\) after separation.
This phenomenon is crucial to solving problems related to charge redistribution. It showcases how charge flows until an equilibrium is reached, reflecting how nature seeks balance.

Force Calculation

Calculating the electrostatic force between two charged objects, such as conducting spheres, relies on Coulomb's Law. This law states that the magnitude of the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them:

• Formula: \( F = k \, rac{Q_1 Q_2}{d^2} \)
• Where \( k \) is Coulomb's constant \( (9 \times 10^9 \, ext{Nm}^2/ ext{C}^2) \)
• \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges
• \( d \) is the distance between the charges

In this exercise, we first calculated the initial force \( F_1 \) using the original charges \(+1 \, ext{C}\) and \(-2 \, ext{C}\). After the charge redistribution, we recalculated the force \( F_2 \) with the new equal charges of \(-\frac{1}{2} \, ext{C}\) on each sphere. Understanding force calculation is essential to predict how charged objects interact in electrostatics.

Conducting Spheres

Conducting spheres are materials that allow electric charges to move freely across their surface. They are significant in understanding charge distribution and force calculation in electrostatics. When we bring two conducting spheres into contact, their charges readily adjust to balance across both spheres. This property makes conductors unique as compared to insulators which hold charges fixed at their locations.
In our problem, we found that upon contact, charge equalizes between the two spheres. This leads to a redistribution of charge that directly affects the electrostatic forces they experience, making the system dynamic. Conducting spheres are used extensively in physics experiments and real-world applications where transferring or balancing charge is necessary.

Electrostatics Problem

Electrostatics problems, like the one we're tackling, explore the effects of electric charges at rest. They involve understanding how charges interact through electric forces based on principles such as Coulomb's Law.
A proper grasp of electrostatics requires knowing how charges influence each other, which forces they exert, and how materials like conductors affect charge distribution. This particular problem provided a practical example of these concepts: from calculating forces between charged spheres to understanding how they redistribute charge upon contact.
Solving electrostatics problems involves a sequential approach, where you:

• Identify known and unknown quantities (charges and distances)
• Apply key physics principles (Coulomb's Law)
• Solve for relationships or quantities (such as force ratios)

This process aids in developing problem-solving skills useful in various scientific contexts.