Question

Three \(-9-\mathrm{mC}\) point charges are located at (0,0) \((3 \mathrm{~m}, 3 \mathrm{~m})\), and \((3 \mathrm{~m},-3 \mathrm{~m})\). What is the magnitude of the electric field at \((3 \mathrm{~m}, 0) ?\) a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) b) \(1.2 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) c) \(1.8 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) d) \(2.4 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) e) \(3.6 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) f) \(5.4 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) g) \(10.8 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\)

Short answer

a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) b) \(2.7 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) c) \(9 \cdot 10^{6} \mathrm{~N} / \mathrm{C}\) d) \(0\mathrm{~N} / \mathrm{C}\) Answer: a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\)

Step-by-step solution

Recall the formula for electric field due to a point charge

The electric field E due to a point charge Q at a distance r from the charge is given by the formula: \[ E = \frac{kQ}{r^2} \] Where k is the electrostatic constant, approximately \(8.99 \cdot 10^{9} \frac{\mathrm{N} \cdot \mathrm{m^2}}{\mathrm{C^2}}\).

Calculate the distance from each charge to the point of interest

First, let's find the distance from each of the charges to the point (3 m, 0), by using the distance formula in two-dimensional space: \[r = \sqrt{(x-x_0)^2 + (y-y_0)^2}\] For the charge at (0, 0): \[r_1 = \sqrt{(3-0)^2 + (0-0)^2} = 3\text{ m}\] For the charge at (3 m, 3 m): \[r_2 = \sqrt{(3-3)^2 + (0-3)^2} = 3\text{ m}\] For the charge at (3 m, -3 m): \[r_3 = \sqrt{(3-3)^2 + (0+3)^2} = 3\text{ m}\]

Calculate the electric field for each point charge

Now, we will use the electric field formula and the calculated distances to find the electric field due to each charge at the point (3 m, 0). For the charge at (0, 0): \[E_1 = \frac{k(-9 \cdot 10^{-9} \text{C})}{(3\text{ m})^2} = -9\cdot10^6\text{ N/C along the x-axis}\] For the charge at (3 m, 3 m): \[E_2 = \frac{k(-9 \cdot 10^{-9} \text{C})}{(3\text{ m})^2} = -9\cdot10^6\text{ N/C along the -y-axis (downward)}\] For the charge at (3 m, -3 m): \[E_3 = \frac{k(-9 \cdot 10^{-9} \text{C})}{(3\text{ m})^2} = -9\cdot10^6\text{ N/C along the y-axis (upward)}\]

Find the vector sum of the electric fields

Next, we need to find the vector sum of these electric fields. Since \(E_2\) and \(E_3\) are along -y and y axes respectively with equal magnitudes, they will cancel each other out. Thus, the total electric field is along the x-axis and has a magnitude of \(-9 \cdot 10^6\text{ N/C}\) (which is the strength of \(E_1\))

Determine the magnitude of the total electric field

Since the electric field is entirely along the x-axis, the magnitude will be the same as its x-component (ignoring the negative sign): \[E_{total} = |-9 \cdot 10^6\text{ N/C}| = 9 \cdot 10^6\text{ N/C}\] After comparing this result with the given choices, the correct answer is: a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\).

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle that describes the electrostatic interaction between electrically charged particles. It fundamentally states how the electric force between two point charges depends on the magnitude of their charges and the distance separating them. The law is mathematically expressed as:\[ F = k\frac{|Q_1Q_2|}{r^2} \]where:

• \( F \) is the force between the charges,
• \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges,
• \( r \) is the distance between the centers of the two charges, and
• \( k \) is the electrostatic constant (\(8.99 \cdot 10^9 \frac{\mathrm{N} \cdot \mathrm{m^2}}{\mathrm{C^2}}\)).

It is important to note the inverse square relationship. This means that as you increase the distance \( r \), the force quickly decreases. This key relationship allows us to understand how charges influence each other over distance.

Electric Field Due to Point Charge

The electric field created by a point charge is a vector field around the charge that represents the force a positive test charge would experience at any given point in the space surrounding the charge. This is significant, as it allows us to understand how various charges affect the space around them.

For a point charge \( Q \), the electric field \( E \) at a distance \( r \) from the charge is given by the formula:\[ E = \frac{kQ}{r^2} \]Where:

• \( E \) is the electric field strength,
• \( Q \) is the charge,
• \( r \) is the distance from the charge, and
• \( k \) is the electrostatic constant.

An important property of electric fields is their direction. For a positive point charge, the field radiates outward, while for a negative point charge, it points inward. This vector nature of electric fields is crucial when multiple charges are present, as interactions may occur simultaneously from different directions.

Vector Addition of Electric Fields

When dealing with electric fields created by multiple point charges, the principle of superposition is applied. This principle states that multiple fields can be added together as vectors to determine the resultant electric field at a certain point. Each electric field is calculated independently and then summed vectorially.
Here's how it works:

• Calculate the electric field due to each charge at the point of interest.
• Use vector addition to sum these individual fields.
• Consider both the magnitude and direction of each field.

In the example given, the electric fields were vectorially combined. Charges at points mirror each other's position across axes create equal and opposite electric fields that cancel out, simplifying the solution process. This is why understanding and using vector addition is critical in physics, as it allows for a complete picture of how multiple factors influence a particular point in space.