Problem 10 - Master Resource Book in JEE Main Physics

Question

There are two charged identical metal spheres \(A\) and \(B\) repel each other with a force \(3 \times 10^{-5} \mathrm{~N}\). Another identical uncharged sphere \(C\) is touched with \(A\) and then placed at the mid-point between \(A\) and \(B\). Net force on \(C\) is (a) \(1 \times 10^{-5} \mathrm{~N}\) (b) \(2 \times 10^{-5} \mathrm{~N}\) (c) \(1.5 \times 10^{-5} \mathrm{~N}\) (d) \(3 \times 10^{-5} \mathrm{~N}\)

Step-by-Step Solution

Answer

(c) \(1.5 \times 10^{-5} \mathrm{~N}\)

Step 1: Understanding Initial Conditions

Initially, spheres \(A\) and \(B\) repel each other with a force of \(3 \times 10^{-5} \text{ N}\). This indicates that the spheres have equal but opposite charges, \( q_A \) and \( q_B \). Because \(A\) and \(B\) are identical and repel each other, we have that \(q_A = q_B \).

Step 2: Charging Sphere C

Sphere \(C\) is initially uncharged. When \(C\) is touched to sphere \(A\), charge will redistribute equally because they are identical metal spheres, each taking half of the total charge. After touching, each sphere \(A\) and \(C\) will have a charge of \( \frac{q_A}{2} \). Since initially \(A\) had charge \( q_A \), after touching, both \(A\) and \(C\) now have \( \frac{q_A}{2} \).

Step 3: Calculating New Forces on Sphere C

After sphere \(C\) is placed at the midpoint between \(A\) and \(B\), let's calculate the forces. Sphere \(A\) exerts a force \(F_{AC}\) on \(C\) given by Coulomb’s law: \[ F_{AC} = k \cdot \frac{\left(\frac{q_A}{2}\right) \left( \frac{q_A}{2} \right)}{(0.5d)^2} \] Similarly, sphere \(B\) exerts a force \(F_{BC}\) on \(C\) because \(B\)'s charge remains \(q_B = q_A\): \[ F_{BC} = k \cdot \frac{q_A \cdot \left( \frac{q_A}{2} \right)}{(0.5d)^2} \] Because \(A\) and \(B\) are equidistant from \(C\) and since \(F_{BC} > F_{AC}\), net force \(F_C = \lvert F_{BC} - F_{AC} \lvert = \frac{1}{2} F_{AB}\).

Step 4: Calculating Net Force

Plug the given force value \( F_{AB} = 3 \times 10^{-5} \text{ N} \) into \(F_C = \frac{1}{2} F_{AB}\): \[ F_C = \frac{1}{2} \times \left(3 \times 10^{-5} \text{ N}\right) = 1.5 \times 10^{-5} \text{ N}\] So, the net force on sphere \(C\) is \(1.5 \times 10^{-5} \text{ N}\).

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental in understanding how charged objects interact. It defines the force of attraction or repulsion between two charged bodies. This force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is expressed as: \[ F = k \cdot \frac{q_1 \cdot q_2}{r^2} \]where:

• F is the force between the charges.
• k is Coulomb's constant \( (8.9875 \times 10^9 \text{ Nm}^2/\text{C}^2) \).
• q_1 and q_2 are the magnitudes of the charges.
• r is the distance between the centers of the two charges.

This law helps predict how charges will behave in various situations. Such as the initial problem statement where two metal spheres, each carrying charge, repel due to the like charges exerting a force on one another.

Charge Distribution

The concept of charge distribution becomes essential when analyzing the touching of sphere A with sphere C. When a charged sphere touches an identical uncharged sphere, the charge distributes evenly. This is because the spheres are conductive and share equal potential. Initially, sphere A had charge \(q_A\). Upon touching sphere C, the charge equally divides, giving both spheres a charge of \(\frac{q_A}{2}\).

• After touching, both have \(\frac{q_A}{2}\) each because they are identical.
• Equal distribution occurs due to identical size and material.

Charge distribution is a crucial concept in electrostatics. It illustrates how proximity affects electrical equilibrium and prepares us for further force calculations in such scenarios.

Force Calculation

Force calculation is important to determine the net force acting on an object in an electric field. After sphere C was charged and placed at the midpoint between spheres A and B, we used Coulomb’s Law to find individual forces.The force of sphere A on C \((F_{AC})\) and sphere B on C \((F_{BC})\) are calculated using:\[ F = k \cdot \frac{q \cdot q}{d^2} \]Given the charges and position:

• A provides a reduced repelling force (half the charge), so \(F_{AC}\) is less.
• B, being at full charge, allows \(F_{BC}\) to be stronger.

The net force is the difference \(F_C = |F_{BC} - F_{AC} |\). Using the formula:\[ F_C = \frac{1}{2} F_{AB} = 1.5 \times 10^{-5} \text{ N} \]This calculation effectively demonstrates how proximity and charge magnitude impact force magnitude.