Problem 64 - Physics

Question

Concept Questions A positive point charge is surrounded by an equipotential surface \(A\), which has a radius of \(r_{A}\). A positive test charge moves from surface \(A\) to another equipotential surface \(B,\) which has a radius of \(r_{B} .\) In the process, the electric force does negative work. (a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge? (b) Is \(r_{B}\) greater than or less than \(r_{A}\) ? Explain your answers. The positive point charge is \(q=+7.2 \times 10^{-8} \mathrm{C},\) and the test charge is \(q_{0}=+4.5 \times 10^{-11} \mathrm{C} .\) The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9} \mathrm{~J}\). The radius of surface \(A\) is \(r_{A}=1.8 \mathrm{~m} .\) Find \(r_{B}\) Check to see that your answer is consistent with your answers to the Concept Questions.

Step-by-Step Solution

Answer

(a) Opposite direction. (b) \(r_B > r_A\). \(r_B \approx 2.7\, \text{m}\).

Step 1: Understanding Force Direction

Since the work done by the electric force is negative, this indicates that the force and the displacement of the test charge are in opposite directions. This means that the electric force is trying to pull the test charge back towards the positive point charge as it moves away.

Step 2: Determine Relation Between rA and rB

Given that work is negative, the test charge is moving against the electric field. In a radial field surrounding a positive charge, a test charge moving outward experiences a force pulling in, so the electric force does negative work, indicating that the test charge moves to a potential surface further away: \(r_B > r_A\).

Step 3: Calculate rB Using Electric Potential

The electric potential at a distance \(r\) from a point charge \(q\) is \(V(r) = \frac{kq}{r}\), where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\). The work done \(W_{AB}\) by an electric field when moving a charge \(q_0\) between two equipotential surfaces is given by \(W_{AB} = q_0 (V_B - V_A)\). Substitute potentials: \[W_{AB} = q_0 \left( \frac{kq}{r_B} - \frac{kq}{r_A} \right)\].

Step 4: Solve for rB

Rearrange the equation from Step 3 to solve for \(r_B\). Given: \(W_{AB} = -8.1 \times 10^{-9} \, \text{J}\), \(q = 7.2 \times 10^{-8} \, \text{C}\), \(q_0 = 4.5 \times 10^{-11} \, \text{C}\), \(r_A = 1.8 \, ext{m}\): \[-8.1 \times 10^{-9} = 4.5 \times 10^{-11} \left( \frac{8.99 \times 10^9 \times 7.2 \times 10^{-8}}{r_B} - \frac{8.99 \times 10^9 \times 7.2 \times 10^{-8}}{1.8} \right)\]Solving gives: \[r_B \approx 2.7 \, ext{m}\].

Step 5: Verify Consistency

Verify that \(r_B > r_A\) (\(2.7 \, ext{m} > 1.8 \, ext{m}\)), consistent with our answers to the concept questions that the test charge moved to a surface further from the point charge because the electric force did negative work.

Key Concepts

Equipotential Surfaces

Equipotential surfaces are imaginary surfaces where every point shares the same electric potential. They are like contour lines on a map, indicating heights, but here they represent potential energy levels around a charge. When dealing with a point charge, these surfaces are spheres centered on the charge.
What makes these surfaces interesting is that it takes no work to move a test charge along one of these surfaces because the potential energy does not change. However, if a charge moves between different equipotential surfaces, work is done either by or against electric forces, depending on the direction relative to the field. This concept is key when analyzing scenarios in electrostatics, such as determining the impact on work and potential when moving between surfaces A and B in the exercise.

Electric Force

The electric force is a fundamental force that acts between charged particles. It follows Coulomb's law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. This force can be either attractive or repulsive:

• Like charges repel each other.
• Opposite charges attract each other.

In our exercise, the electric force acts on the test charge as it moves from surface A to surface B. The nature of the force means that it acts in the opposite direction to the displacement, indicating that negative work is done. Here, the electric force tries to 'pull' the test charge back towards the positive point charge, demonstrating the nature of electric repulsion in this context.

Work and Energy

Work and energy in the context of electric forces can be a bit tricky. When a charge moves in an electric field, work is done, which can either increase or decrease its potential energy. Work done by an electric force is calculated using the formula:
\[ W = q(V_B - V_A) \] \(W\) is the work done by the force, \(q\) is the charge moving between points B and A, and \(V_B\) and \(V_A\) are the electric potentials at those points.
In cases where the work done is negative, as in our exercise, it means that energy is being taken out of the charge's potential energy, which aligns with the fact that the force is opposed to displacement. The electric field attempts to pull the test charge back, acting as if it were performing 'negative work' against the motion.

Charge Interaction

Charge interaction is a cornerstone of electrostatics, describing how charged particles influence each other. In this scenario, the interaction between a positive point charge and a positive test charge is central. Understanding charge interaction involves recognizing two things:

• The force type - repulsion or attraction based on charge signs.
• The radius of interaction in the case of point charges, or the distance between them.

Here, both the charges involved are positive, so they repel each other.This interaction typically results in the test charge feeling a force that pushes it away from the point charge. This is why when work is performed to move a charge further away (from surface A to B), it's moving against the interactive force. Thus, negative work indicates the test charge overcomes repulsion to go to a surface with a higher potential radius as shown by our calculation of the new radius \(r_B\).