Problem 64 - Physics

Question

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. Problem 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

Electric force opposes displacement; hence, \( r_B < r_A \), specifically \( r_B \approx 1.6 \mathrm{~m} \).

Step 1: Understanding Equipotential Surfaces

Equipotential surfaces are surfaces where the electric potential is the same at all points. When moving along an equipotential surface, no work is done by the electric force because the potential change is zero. Therefore, if work is done in moving from surface A to surface B, the potential must differ between the two surfaces.

Step 2: Analyzing Direction of Electric Force

Since the electric force does negative work when the test charge moves from surface A to surface B, the force acts opposite to the direction of displacement. This implies the test charge is moving against the field lines emanating from the positive point charge.

Step 3: Relationship Between Equipotential Surfaces and Radii

If the test charge moves against the electric field of the positive point charge, it is moving to a surface at higher electrical potential. For a positive point charge, as the radius increases, potential decreases. Therefore, to increase potential, it must move to a smaller radius. Thus, it indicates that \( r_B < r_A \).

Step 4: Determine Potential Difference

The work done by the electric force can be related to the potential difference between the two surfaces. The work done \( W \) is related to the potential difference \( V_A - V_B \) by the formula: \( W = q_0 (V_A - V_B) \). Given that \( W_{AB} = -8.1 \times 10^{-9} \mathrm{~J} \) and \( q_0 = +4.5 \times 10^{-11} \, \mathrm{C} \), we can find the potential difference.

Step 5: Calculating Potential at Surfaces A and B

The electric potential \( V \) at a distance \( r \) from a point charge \( q \) is given by: \( V = \frac{kq}{r} \). For surface A: \( V_A = \frac{kq}{r_A} \). Let's use this to find the potential difference: \[ V_A - V_B = \frac{kq}{r_A} - \frac{kq}{r_B} \].

Step 6: Solving for \( r_B \)

Using the expression for potential difference: \( \frac{kq}{r_A} - \frac{kq}{r_B} = \frac{W_{AB}}{q_0} \). Rearranging gives \( \frac{1}{r_B} = \frac{1}{r_A} - \frac{W_{AB}}{q_0 kq} \). With known \( W_{AB} = -8.1 \times 10^{-9} \mathrm{~J} \), \( r_A = 1.8 \mathrm{~m} \), \( k = 8.99 \times 10^9 \mathrm{~N \cdot m^2/C^2} \), we solve for \( r_B \).

Step 7: Calculate and Verify \( r_B \)

Substitute the values: \( \frac{k \times 7.2 \times 10^{-8}}{1.8} - \left( - \frac{8.1 \times 10^{-9}}{4.5 \times 10^{-11}} \right) = \frac{k \times 7.2 \times 10^{-8}}{r_B} \). Solving the equation gives \( r_B \approx 1.6 \mathrm{~m} \). This is consistent with our conceptual understanding that \( r_B < r_A \).

Key Concepts

Equipotential Surfaces

Equipotential surfaces are quite fascinating when it comes to electric potential. Imagine them like invisible blankets wrapping around a charge where the electric potential (or energy per unit charge) is constant.
Such surfaces make understanding electric fields manageable because if you move a test charge along this blanket,

• there's no work done by the electric field,
• because the charge does not "feel" any difference in potential energy.

But the moment the charge steps off this surface onto another equipotential surface, the game changes. This change signifies a difference in potential energy and, hence, some work will be done either by or against the electric field. Thus, in simplest terms, these equipotential shells tell us where a charge experiences no change in potential energy.

Positive Point Charge

A positive point charge plays an essential role in the study of electric fields. Picture it as a tiny center of electric force, radiating field lines all around it like sun rays. These field lines point outward, indicating the direction that a positive test charge would naturally want to move if placed in the field.
This orientation of field lines is crucial when considering problems involving electric potential surfaces.
With a positive point charge, as one moves further away:

• the field becomes weaker (as the lines are spread out over a larger area),
• the electric potential decreases.

Remembering these principles can help understand how and why a test charge moves in space surrounding a point charge.

Electric Force

Electric force is powerful and ubiquitous in its action over charged objects. Fundamentally, it is the push or pull experienced by a charge due to the presence of electric fields.
In this scenario, the test charge moving from one equipotential surface to another experiences this electric force. The direction of the electric force here is determined by the electric field. The force acts in a direction that aligns with the field lines if the charge is positive. But if the work done by this force is negative, as in the given exercise, it indicates that the test charge moves opposite to the direction of the electric force.
This is akin to moving an object uphill against gravity, where the force on the object wants it to move back. It's important to note that a negative work done by electric force signifies resistance against the field's natural direction.

Work Done by Electric Field

Work done by the electric field is an essential concept that connects electric force and potential energy. When we talk about moving a test charge between two points, or surfaces, in an electric field, we often ask how much work is required or done. This work is determined by the change in electric potential energy between the points.
If you remember the expression

• \( W = q_0 (V_A - V_B) \),

it essentially tells us that work is the product of the test charge and the potential difference it experiences.In the given exercise, the work done by the electric field is negative, \(-8.1 \times 10^{-9} \mathrm{~J}\), signifying the field is doing work against the motion of the charge, similar to how pushing against resistance requires a force. This negative aspect of work illustrates moving to a higher potential area around a positive charge, where the electric force naturally resists such movement.