Problem 75 - University Physics with Modern Physics

Question

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere \(A\) has a radius three times that of sphere \(B\). Let \(Q_A\) and \(Q_B\) be the charges on the two spheres, and let \(E_A\) and \(E_B\) be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio \(Q_B/Q_A\) and (b) the ratio \(E_B/E_A\)?

Step-by-Step Solution

Answer

(a) \(\frac{1}{3}\), (b) 3.

Step 1: Identify Given Variables and Concepts

We have two spheres, A and B, with sphere A having a radius three times that of sphere B. The two spheres have equal electric potential at their surfaces. The charge on sphere A is \(Q_A\) and on sphere B is \(Q_B\). We need to determine the ratio of their charges \(Q_B/Q_A\) and the ratio of the electric field magnitudes at their surfaces \(E_B/E_A\).

Step 2: Apply Equal Electric Potential Condition

The electric potential \(V\) on the surface of a sphere is given by \(V = \frac{kQ}{r}\), where \(Q\) is the charge, \(r\) is the radius, and \(k\) is Coulomb's constant. For sphere A, \(V_A = \frac{kQ_A}{3r}\) and for sphere B, \(V_B = \frac{kQ_B}{r}\). Since the potentials are equal, \(\frac{kQ_A}{3r} = \frac{kQ_B}{r}\).

Step 3: Solve for Ratio of Charges

From \(\frac{kQ_A}{3r} = \frac{kQ_B}{r}\), the \(k\) and \(r\) terms cancel out, leading to \(\frac{Q_A}{3} = Q_B\). Thus, \(Q_B = \frac{Q_A}{3}\), resulting in the ratio \(Q_B/Q_A = \frac{1}{3}\).

Step 4: Determine Electric Field Magnitude Relation

The electric field \(E\) on the surface of a sphere is given by \(E = \frac{kQ}{r^2}\). For sphere A, \(E_A = \frac{kQ_A}{(3r)^2}\), and for sphere B, \(E_B = \frac{kQ_B}{r^2}\).

Step 5: Solve for Ratio of Electric Field Magnitudes

Since \(Q_B = \frac{Q_A}{3}\), we substitute this into \(E_B = \frac{kQ_B}{r^2}\) giving \(E_B = \frac{k\left(\frac{Q_A}{3}\right)}{r^2} = \frac{kQ_A}{3r^2}\). Compare \(E_B\) and \(E_A\), we have \(E_A = \frac{kQ_A}{9r^2}\). Thus, \(E_B/E_A = \frac{3}{1} = 3\).

Key Concepts

Charge Ratio

When considering two charged spheres, it is important to understand how the charges are distributed relative to each other, particularly when their surfaces have the same electric potential. In this exercise, you have Sphere A and Sphere B. Sphere A has a radius that is three times larger than that of Sphere B. According to the problem, both spheres have equal electric potentials at their surfaces.

The electric potential ( \( V \) ) of a sphere is given by the formula \( V = \frac{kQ}{r} \), where \( Q \) is the charge, \( r \) is the radius, and \( k \) is Coulomb's constant. Since both spheres have the same potential, we set their equations equal and cancel out the appropriate terms:

• Sphere A: \( \frac{kQ_A}{3r} \)
• Sphere B: \( \frac{kQ_B}{r} \)

By setting \( \frac{kQ_A}{3r} = \frac{kQ_B}{r} \), the \( k \) and \( r \) terms cancel, leaving \( \frac{Q_A}{3} = Q_B \).

Thus, the ratio of charges \( \frac{Q_B}{Q_A} \) becomes \( \frac{1}{3} \). This means that the charge on Sphere B is one-third of the charge on Sphere A. Understanding this relation helps in predicting how different the spheres behave under similar electric potential conditions.

Electric Field Magnitude

The electric field magnitude of a sphere is another crucial concept, especially when comparing two spheres of different sizes but under similar electric potential conditions. For each sphere, the electric field (\( E \)) at the surface is derived by the formula: \( E = \frac{kQ}{r^2} \), where \( Q \) is the charge, \( r \) is the radius, and \( k \) is Coulomb's constant.

For Sphere A, which has three times the radius of Sphere B, the electric field is represented as follows:

• Sphere A: \( E_A = \frac{kQ_A}{(3r)^2} = \frac{kQ_A}{9r^2} \)
• Sphere B: \( E_B = \frac{kQ_B}{r^2} \)

With the earlier determined charge relationship \( Q_B = \frac{Q_A}{3} \), substitute this in for Sphere B: \( E_B = \frac{k(\frac{Q_A}{3})}{r^2} = \frac{kQ_A}{3r^2} \).

When you compare \( E_B \) to \( E_A \), you'll find that \( \frac{E_B}{E_A} = \frac{3}{1} \), which indicates that the electric field on the surface of Sphere B is three times that of Sphere A. This outcome illustrates how smaller spheres can have larger electric fields even when they hold less overall charge.

Coulomb's Law

Coulomb's Law is fundamental in understanding the forces and electric potential in charged objects. It states that the electric force (\( F \)) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them: \[ F = k \frac{|Q_1 Q_2|}{r^2} \].

In the context of the two spheres discussed in this exercise, Coulomb's Law helps to understand why the electric fields and potential change with the variation of radius and charge. The constant \( k \) here represents Coulomb's constant, a value that helps to scale the force and potential energy involved.

When spheres are involved, assessing the electric potential or field requires adjustments because we're often not just interested in forces between point charges, but in how fields manifest around surfaces. In this exercise, it helped show the equal potentials on spheres despite different sizes, and how the relation of distances and radii play a more pronounced role due to the radial distribution of charge.

Thus, Coulomb's Law is a keystone for bridging the relationship between potential, field magnitudes, and charge ratios, enabling a deeper comprehension of electrostatic interactions.