Problem 50 - Master Resource Book in JEE Main Physics

Question

Two metallic spheres \(A\) and \(B\) of same radii one solid and other hollow are charged to the same potential. Which of the two will hold more charge? (a) Sphere \(A\) (b) Sphere \(B\) (c) Both spheres (d) None of these

Step-by-Step Solution

Answer

Both spheres hold the same amount of charge. (c) Both spheres.

Step 1: Understand the Concept

The potential (V) of a charged sphere is given by the formula \( V = \frac{kQ}{R} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere. Since both spheres have the same potential and radius, we can find which holds more charge based on their structures.

Step 2: Analyze Charge Distribution

For a solid sphere, the charge is distributed throughout the volume, while for a hollow sphere, the charge resides on the outer surface. However, the potential depends only on the total charge and radius for each sphere, not how it is distributed.

Step 3: Apply the Formula for Each Sphere

Since each sphere has the same potential \( V \), we set the formulas equal: \( \frac{kQ_A}{R} = \frac{kQ_B}{R} \) where \( Q_A \) and \( Q_B \) are charges of sphere A and B. Given radii and potentials are identical, \( Q_A = Q_B = Q \).

Step 4: Conclusion

Since \( Q_A = Q_B \), both spheres hold the same total charge, regardless of being solid or hollow. Therefore, the answer is that both spheres can hold the same amount of charge.

Key Concepts

Charged Spheres

When deep diving into the topic of charged spheres, it's crucial to understand how their physical nature affects their charge capacity. Metallic spheres like the ones mentioned, whether solid or hollow, can be charged up to a certain potential. This potential is the electric potential energy per unit charge at the sphere’s surface. Both spheres in the problem are metallic, meaning they are conductive and allow the charge to distribute itself evenly in response to the electric field.

• Solid Sphere: Charge is not only distributed on the surface but also throughout the entire volume of the conductor.
• Hollow Sphere: Charge resides only on the surface of the conductor.

It's important to note that for the purpose of calculating potential and charge, it is the overall charge on the sphere that matters, not how it is spread across the sphere.

Potential Formula

The potential formula is key when determining how much charge a sphere can hold. The potential V of a sphere is given by the formula:\[V = \frac{kQ}{R}\]where:

• \(V\) is the electric potential.
• \(k\) is Coulomb's constant, a value of approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).
• \(Q\) is the charge on the sphere.
• \(R\) is the radius of the sphere.

Since the potential and radius are constant in our problem, both spheres can logically hold the same charge regardless of their physical structure.

Charge Distribution

Charge distribution in spheres affects how we perceive the placement of charge. In electrostatics, for objects like spheres, the distribution doesn't influence the potential, but rather the field outside the sphere. The distribution:

• Solid Sphere: Charges repel and attempt to move as far apart as possible due to Coulomb’s law, resulting in a uniform distribution throughout the sphere.
• Hollow Sphere: All the charge resides on the surface. Even with no internal volume charges, the conductive nature channels them outward.

Hence, for calculating the potential, as seen with the potential formula, distribution within a solid or hollow sphere does not change the result.

Coulomb's Constant

Coulomb's constant \(k\) is a fundamental principle in electrostatics, helping us quantify force and potential. This constant arises in many key formulas:\[k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\]In the context of our problem:

• It helps relate the potential and charge in the formula \(V = \frac{kQ}{R}\).
• Represents the strength of the electric force in vacuum meaning it’s the degree of force a unit charge experiences in space.
• Ensures that calculations of electrostatic potentials and fields adhere to standardized physical laws.

By understanding and applying Coulomb's constant, we tie together the variables of charge, potential, and distance efficiently.