Problem 81 - 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) \( \frac{Q_B}{Q_A} = \frac{1}{3} \). (b) \( \frac{E_B}{E_A} = 3 \).

Step 1: Understanding Electric Potential on Spheres

The electric potential \( V \) at the surface 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 the electric potential is the same for both spheres, we have \( \frac{kQ_A}{R_A} = \frac{kQ_B}{R_B} \).

Step 2: Relating Radii of the Spheres

We know sphere A's radius \( R_A \) is three times that of sphere B or \( R_A = 3R_B \). Substituting this into the potential equation gives \( \frac{Q_A}{3R_B} = \frac{Q_B}{R_B} \).

Step 3: Solving for Charge Ratio

We equate the expressions for potential and solve for \( Q_B/Q_A \). Simplifying \( \frac{Q_A}{3} = Q_B \) gives the charge ratio \( \frac{Q_B}{Q_A} = \frac{1}{3} \).

Step 4: Understanding Electric Field on a Sphere

The electric field \( E \) at the surface of a charged sphere is given by \( E = \frac{kQ}{R^2} \). We need to find \( E_B / E_A \) using this formula for each sphere.

Step 5: Calculating the Electric Field Ratio

Substitute \( Q_B = \frac{Q_A}{3} \) and \( R_A = 3R_B \) into the electric field formula: \( E_B = \frac{kQ_B}{R_B^2} = \frac{k(\frac{Q_A}{3})}{R_B^2} \) and \( E_A = \frac{kQ_A}{(3R_B)^2} = \frac{kQ_A}{9R_B^2} \). Thus, \( \frac{E_B}{E_A} = \frac{\frac{k \cdot (\frac{Q_A}{3})}{R_B^2}}{\frac{kQ_A}{9R_B^2}} \).

Step 6: Simplifying the Electric Field Ratio

Cancelling terms, the electric field ratio \( \frac{E_B}{E_A} = \frac{9}{3} = 3 \), meaning the electric field on sphere B is three times that on sphere A.

Key Concepts

Charged Spheres

Metal spheres can become electrically charged by either gaining or losing electrons. This causes a build-up of electric charge, leading to an electric potential on their surfaces. When two spheres have identical electric potentials, the potential due to the charge and the radius must be balanced. The electric potential, given by \( V = \frac{kQ}{R} \), depends on the charge \( Q \) and the radius \( R \) of each sphere. In our exercise, we have two spheres with the same potential but different sizes. Sphere \( A \) has a radius that is three times larger than that of sphere \( B \).
This difference in radius means that to maintain the same potential on both spheres, their charges must also differ. By substituting the equation for electric potential, we can explore the relationship between their charges and sizes.

Electric Field

An electric field represents the force per unit charge exerted by a charge. For charged spheres, the electric field at any surface point can be determined using the formula \( E = \frac{kQ}{R^2} \). This formula shows that the electric field not only depends on the charge \( Q \) but also inversely on the square of the radius \( R \).
In our scenario, even though sphere \( A \) and sphere \( B \) have the same electric potential, their electric fields can be very different due to the difference in radii. The smaller sphere \( B \) generates a stronger electric field than sphere \( A \) because the electric field concentration is greater when the radius is smaller. This is because the same amount of charge is spread over a smaller surface area.

Coulomb's Law

Coulomb's law is a fundamental principle that describes the interaction between electric charges. Defined using the formula \( F = \frac{kQ_1Q_2}{r^2} \), it helps in understanding the electric force between two point charges. Here, \( k \) is Coulomb's constant, \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
In the case of two charged spheres, we can apply concepts similar to Coulomb’s law to understand their interactions and calculate electric potential. Coulomb's law illustrates how charges interact with one another and manifest as electric forces and fields. This is directly linked to calculating fields and potentials on charged spheres, reinforcing the observed differences in electric fields between spheres of different sizes.