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Chapter 17: Problem 50

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

Short Answer

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

Step by step solution

01

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.
02

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.
03

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, \( Q_A = Q_B = Q \).
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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鈥檚 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鈥檚 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.

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Most popular questions from this chapter

A charged body has an electric flux \(\phi\) associated with it. The body is now placed inside a metallic container. The electric flux, \(\phi_{1}\) associated with the container will be (a) \(\phi_{1}=0\) (b) \(0<\phi_{1}<\phi\) (c) \(\phi_{1}=\phi\) (d) \(\phi_{1} \geq \phi\)

Two free protons are separated by a distance of \(1 \AA\). If one proton is kept at least distance and the other is released, the kinetic energy of second proton when it is at infinite separation is (a) \(23.0 \times 10^{-19} \mathrm{~J}\) (b) \(11.5 \times 10^{-19} \mathrm{~J}\) (c) \(2.3 \times 10^{-19} \mathrm{~J}\) (d) zero

A sphere of radius \(r\) is charged to a potential \(V\). The outward pull per unit area of its surface is given by (a) \(\frac{4 \pi \varepsilon_{0} V^{2}}{r^{2}}\) (b) \(\frac{\varepsilon_{0} V^{2}}{2 r^{2}}\) (c) \(\frac{2 \pi \varepsilon_{0} V^{2}}{r^{2}}\) (d) \(\frac{\varepsilon_{0} V^{2}}{4 r^{2}}\)

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The force on each plate of parallel plate capacitor has a magnitude equal to \(\frac{1}{2} Q E\), where \(Q\) is the charge on the capacitor and \(E\) is the magnitude of electric field between the plates. Then (a) \(\frac{E}{2}\) contributes to the force against which the plates are moved (b) \(\frac{E}{3}\) contributes to the force against which the plates are moved (c) \(E\) contributes force against which the plates are moved (d) None of the above
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