Problem 26 - College Physics

Question

Two fixed charges, \(-4.0 \mu \mathrm{C}\) and \(-5.0 \mu \mathrm{C},\) are separated by a certain distance. (a) Is the net electric field at a location halfway between the two charges (1) directed toward the \(-4.0 \mu \mathrm{C}\) charge, (2) zero, or (3) directed toward the \(-5.0 \mu \mathrm{C}\) charge? Why? (b) If the charges are separated by \(20 \mathrm{~cm},\) calculate the magnitude of the net electric field halfway between the charges.

Step-by-Step Solution

Answer

(a) The net electric field is directed toward the \(-5.0 \mu C\) charge. (b) The magnitude is \(0.899 \times 10^6 \text{ N/C}\).

Step 1: Identify Given Data

We are provided with two negative charges: \ \( Q_1 = -4.0 \mu C = -4.0 \times 10^{-6} C \) and \( Q_2 = -5.0 \mu C = -5.0 \times 10^{-6} C \). They are separated by a distance of \( d = 20 \text{ cm} = 0.2 \text{ m} \).

Step 2: Understanding the Electric Field Direction

Since both charges are negative, the electric field lines point towards each charge. Halfway between the two charges, the field due to \( Q_1 \) is directed toward \( Q_1 \) and the field due to \( Q_2 \) is directed toward \( Q_2 \). The magnitude of the electric field is stronger from the charge with the larger magnitude, which is \( Q_2 \). Hence, the net electric field will be directed toward the \( -5.0 \mu C \) charge.

Step 3: Calculate Electric Field from Each Charge

The electric field due to a point charge \( Q \) at a distance \( r \) is given by:\[ E = \frac{k |Q|}{r^2} \]Halfway between the charges, the distance from each charge is \( r = 0.1 \text{ m} \). So, for \( Q_1 \):\[ E_1 = \frac{(8.99 \times 10^9 \frac{N \cdot m^2}{C^2}) \cdot |-4.0 \times 10^{-6} C|}{(0.1)^2} \]And for \( Q_2 \):\[ E_2 = \frac{(8.99 \times 10^9 \frac{N \cdot m^2}{C^2}) \cdot |-5.0 \times 10^{-6} C|}{(0.1)^2} \]

Step 4: Compute Electric Fields Individually

Calculate each electric field:\[ E_1 = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6}}{0.01} = 3.596 \times 10^6 \text{ N/C} \]\[ E_2 = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{0.01} = 4.495 \times 10^6 \text{ N/C} \]

Step 5: Determine Net Electric Field Magnitude

The net electric field midway is the difference because the fields are in opposite directions:\[ E_{net} = E_2 - E_1 = 4.495 \times 10^6 - 3.596 \times 10^6 = 0.899 \times 10^6 \text{ N/C} \]The direction is toward the \( -5.0 \mu C \) charge.

Key Concepts

Point Charges

Point charges are fundamental concepts in physics and electric field studies. These are tiny, charged objects with either positive or negative electric load located at a specific point in space. The charges in this exercise, - **Charge 1:** \( Q_1 = -4.0 \mu C \) - **Charge 2:** \( Q_2 = -5.0 \mu C \)- are both negative. Though we often refer to them as points, point charges are really simplifications that help us analyze electric fields without getting bogged down in physical size. These charges create an electric field that can influence other charges placed in the field. Understanding point charges is key to mastering electric field concepts, as they represent localized sources of electric force.

Coulomb's Law

Coulomb's Law describes the force between two point charges. It provides the formula for calculating the electric force acting between them, given as:\[ F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} \]where - **\( F \)** is the magnitude of force,- **\( k \)** is Coulomb's constant \( (8.99 \times 10^9 \frac{N \cdot m^2}{C^2}) \), - **\( Q_1 \)** and **\( Q_2 \)** are the point charges, and- **\( r \)** is the distance between the charges.- In terms of electric fields, Coulomb's law helps in calculating the field produced by a single point charge: \[ E = \frac{k \cdot |Q|}{r^2} \]This derivation emphasizes the inverse square nature of the field, meaning as the distance \( r \) increases, the electric field strength decreases by the square of that distance.

Electric Field Direction

The direction of the electric field created by a point charge depends on the sign of the charge. For positive point charges, the electric field radiates outward from the charge, pushing away. Conversely, for negative charges, the field points inward, toward the charge. In this exercise, because the charges are negative, the electric fields point toward the respective charges:- **Field from \( Q_1 \)** is directed toward \( Q_1 \)- **Field from \( Q_2 \)** is directed toward \( Q_2 \)When observing the net electric field at a point between these charges, you'll notice the stronger field, from the larger \( |Q_2| \), dominates. Thus, the net electric field direction between them is toward the \( -5.0 \mu C \) charge, as its strength is greater than \( -4.0 \mu C \).

Electric Field Magnitude

To find the electric field magnitude at a point, we calculate the field from each charge and then account for their directions. Each field's magnitude is given by \[ E = \frac{k \cdot |Q|}{r^2} \]where \( r \) is the distance from the charge to the point. For two charges \(-4.0 \mu C\) and \(-5.0 \mu C\) separated by \(20 \text{ cm}\), the distance to the midpoint from each charge is \(0.1 \text{ m}\). The field magnitudes are:- **\( E_1 = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6}}{0.01} = 3.596 \times 10^6 \text{ N/C} \)** from \( Q_1 \)- **\( E_2 = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{0.01} = 4.495 \times 10^6 \text{ N/C} \)** from \( Q_2 \)The net field is the difference due to their opposite directions:- \[ E_{net} = E_2 - E_1 = 4.495 \times 10^6 - 3.596 \times 10^6 = 0.899 \times 10^6 \text{ N/C} \]- The net field magnificently shows how individual fields combine considering vector properties.