Problem 17 - University Physics with Modern Physics

Question

Point charges \(q_1 = +\)2.00 \(\mu\)C and \(q_2 = -\)2.00 \(\mu\)C are placed at adjacent corners of a square for which the length of each side is 3.00 cm. Point \(a\) is at the center of the square, and point \(b\) is at the empty corner closest to \(q_2\) . Take the electric potential to be zero at a distance far from both charges. (a) What is the electric potential at point a due to \(q_1\) and \(q_2\)? (b) What is the electric potential at point \(b\)? (c) A point charge \(q_3 = -\)5.00 \(\mu\)C moves from point \(a\) to point \(b\). How much work is done on \(q_3\) by the electric forces exerted by \(q_1\) and \(q_2\)? Is this work positive or negative?

Step-by-Step Solution

Answer

Both potentials are zero. Zero work is done; neither positive nor negative.

Step 1: Determine the Potential at the Center

First, calculate the distance from each charge to the center of the square. Using the Pythagorean theorem, since each side of the square is 3.00 cm, the diagonal of the square is \( s\sqrt{2} = 3.00\sqrt{2} \) cm. The distance from each charge to the center is half of this diagonal: \( \frac{3.00\sqrt{2}}{2} \approx 2.12 \) cm. The electric potential at the center from a charge \( q \) at a distance \( r \) is given by \( V = \frac{kq}{r} \), where \( k = 8.99 \times 10^9 \) N·m²/C². Calculate the potentials due to both charges at this distance and add them.

Step 2: Calculate Potential Due to Each Charge at Point a

For \( q_1 = +2.00 \mu \)C:\[ V_1 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-6}}{0.0212} \approx 8.49 \times 10^5 \text{ V} \]For \( q_2 = -2.00 \mu \)C:\[ V_2 = \frac{8.99 \times 10^9 \times (-2.00 \times 10^{-6})}{0.0212} \approx -8.49 \times 10^5 \text{ V} \]The total potential at point \( a \) is \( V_1 + V_2 = 0 \) V.

Step 3: Determine the Potential at the Corner Point b

The distance from \( q_1 \) at the opposite corner to point \( b \) is the side length of the square, 3.00 cm. The distance from \( q_2 \) to point \( b \) is also 3.00 cm. Calculate the potential due to each charge at point \( b \) and then sum them.For \( q_1 \):\[ V_1b = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-6}}{0.03} \approx 5.99 \times 10^5 \text{ V} \]For \( q_2 \):\[ V_2b = \frac{8.99 \times 10^9 \times (-2.00 \times 10^{-6})}{0.03} \approx -5.99 \times 10^5 \text{ V} \]The total potential at point \( b \) is \( V_1b + V_2b = 0 \) V.

Step 4: Calculate Work Done on Charge Moving from a to b

Work done by electric forces is the change in potential energy: \( W = -\Delta U = -q_3 (V_b - V_a) \) where \( V_b \) and \( V_a \) are potentials at points \( b \) and \( a \), respectively.Since the potential at both points is zero:\[ W = -(-5.00 \times 10^{-6})(0 - 0) = 0 \] Joules.Thus, no work is done on \( q_3 \).

Key Concepts

Point Charge

A point charge is a fundamental concept in electromagnetism that refers to a charged object that is considered to have no physical dimensions, just a specific amount of charge at a theoretical point in space. When dealing with point charges, we often use Coulomb's Law to describe the force between any two charges. The formula is:

• \( F = \frac{k |q_1 \, q_2|}{r^2} \)

where:

• \( F \) is the force between the charges,
• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2) \),
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and
• \( r \) is the distance between the centers of the two charges.

In this scenario, we have two point charges: \( q_1 = +2.00 \, \mu C \) and \( q_2 = -2.00 \, \mu C \), placed at opposite corners of a square. Here, the charges are equal in magnitude but opposite in sign, creating a symmetrical situation that helps simplify potential calculations.When multiple point charges are involved, as in this exercise, the principle of superposition is key. This principle allows us to calculate the potential from each charge separately and then sum these values to obtain the total potential at any given point.

Work Done by Electric Forces

Work done by electric forces involves transferring energy from one form to another by moving a charge in an electric field. The work done is related to the change in electric potential energy of the system. The concept can be explained as follows:

• The work (\( W \)) done by electric forces when a charge \( q_3 \) moves from point \( a \) to point \( b \) in an electric field is given by:
• \( W = -\Delta U = -q_3 (V_b - V_a) \)

Here:

• \( \Delta U \) is the change in electric potential energy,
• \( V_b \) and \( V_a \) are the electric potentials at point \( b \) and point \( a \) respectively, and
• \( q_3 \) is the moving charge.

In our exercise, since the electric potential at both points \( a \) and \( b \) are zero (as calculated in the original solution), substituting these values into the expression shows that:

• \( W = -q_3 (0 - 0) = 0 \) Joules.

This result indicates that no net work is done on charge \( q_3 \), which reinforces that work is only done when there is a change in potential.

Electric Potential Calculation

Electric potential is a concept that measures the potential energy per unit charge at a certain point in an electric field. Calculating the electric potential involves considering each point charge's contribution to the total electric potential at a specific location.The formula for the electric potential (\( V \)) due to a single point charge \( q \) at a distance \( r \) is:

• \( V = \frac{kq}{r} \)

where:

• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2) \),
• \( q \) is the charge, and
• \( r \) is the distance from the charge to the point in question.

When calculating electric potential due to multiple charges, the potential values due to each charge are algebraically added, taking into account the sign of each charge:

• A positive charge contributes a positive potential, and
• A negative charge contributes a negative potential.

In this scenario, at both the center of the square (point \( a \)) and the corner closest to \( q_2 \) (point \( b \)), the potentials from \( q_1 \) and \( q_2 \) cancel each other out.This results in a net electric potential of zero at both points, as shown in the step-by-step calculations, thereby simplifying the work done in moving a charge between these points.