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Chapter 11: Problem 1579

Two charged spheres of radii \(R_{1}\) and \(R_{2}\) having equal surface charge density. The ratio of their potential is ..... (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)\) (B) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

Short Answer

Expert verified
The ratio of their potential is \(\boxed{\text{(D)}\ \left(\frac{R_{1}}{R_{2}}\right)}\).

Step by step solution

01

Understand surface charge density and electric potential.

Surface charge density is given by the amount of charge per unit area (σ), and electric potential (V) of a charged sphere is the amount of work required to bring a unit positive charge from infinity to the point at a certain distance from the sphere. The relationship between them can be given as \(V = \frac{kQ}{r}\), where k is Coulomb's constant, Q is the total charge on the sphere, and r is the distance from the center of the sphere to the point where electric potential is measured.
02

Write down the given values.

Two charged spheres have equal surface charge densities (σ) and radii R1 and R2, respectively. To calculate the total charge on each sphere, use the formula \(Q = σ × A\), where A is the surface area. Since the spheres have equal surface charge densities, their total charges can be represented as \(Q_{1} = σ × 4πR_{1}^{2}\) and \(Q_{2} = σ × 4πR_{2}^{2}\).
03

Write down the electric potentials for both spheres.

Use the formula \(V = \frac{kQ}{r}\) to write down the electric potentials for both spheres, keeping in mind that the potentials are measured at their surfaces. Therefore, the distances r in the formulas are equal to the radii of the spheres: \(V_{1} = \frac{kQ_{1}}{R_{1}}\) \(V_{2} = \frac{kQ_{2}}{R_{2}}\)
04

Substituting the total charge expressions and finding the ratio of the potentials.

Substitute the total charge expressions from step 2 in the potential formulas: \(V_{1} =\frac{k(σ × 4πR_{1}^{2})}{R_{1}}\) \(V_{2} =\frac{k(σ × 4πR_{2}^{2})}{R_{2}}\) Now, find the ratio of the potentials \(\frac{V_{2}}{V_{1}}\): \(\frac{V_{2}}{V_{1}} = \frac{\frac{k(σ × 4πR_{2}^{2})}{R_{2}}}{\frac{k(σ × 4πR_{1}^{2})}{R_{1}}}\) Simplify the equation and cancel common terms: \(\frac{V_{2}}{V_{1}} = \frac{\frac{(σ × 4πR_{2}^{2})}{R_{2}}}{\frac{(σ × 4πR_{1}^{2})}{R_{1}}} = \frac{R_{1}R_{2}^{2}}{R_{2}R_{1}^{2}}\) \(\frac{V_{2}}{V_{1}} = \frac{R_{1}}{R_{2}}\) The ratio of their potential is: \(\boxed{\text{(D)}\ \left(\frac{R_{1}}{R_{2}}\right)}\)

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