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

Solution: After following the above steps and calculating the individual electric field components and their magnitudes, summing them up, and using the Pythagorean theorem to find the total electric field, we will get the magnitude of the electric field at point (3 m, 0).

Step-by-step solution

Find the Distance Between Charges and Point

The first thing we need to do is find the distances between each of the point charges and the point \((3 \mathrm{~m}, 0)\) where we want to calculate the electric field. We can use the distance formula for this: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ Let's find these distances.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $$d_1 = \sqrt{(3-0)^2+(0-0)^2} = 3 \text{ m}$$ $$d_2 = \sqrt{(3-3)^2+(0-3)^2} = 3 \text{ m}$$ $$d_3 = \sqrt{(3-3)^2+(0+3)^2} = 3 \text{ m}$$

Find the Electric Field Due to Each Charge

Now that we have the distances, let's use Coulomb's Law to find the electric field due to each charge at the point. The electric field produced by a charge q at a distance d is given by the formula: $$E = \frac{k\cdot |q|}{d^2}$$ where \(k=8.9875 \times 10^9 \ \mathrm{Nm^2C^{-2}}\) is the Coulomb's constant, and all charges are \(-9 \ \mathrm{mC}\). Let's calculate the electric fields produced by each charge. $$E_1 = \frac{k\cdot |-9 \times 10^{-3}|}{3^2}$$ $$E_2 = \frac{k\cdot |-9 \times 10^{-3}|}{3^2}$$ $$E_3 = \frac{k\cdot |-9 \times 10^{-3}|}{3^2}$$

Find the Direction of Each Electric Field

The electric field is a vector, so we need to find its direction as well. The electric field produced by a negative charge is always pointing towards the negative charge. Let's visualize the direction of the electric field produced by each charge: 1. \(E_1\): \((-1,0)\) direction (leftwards). 2. \(E_2\): At \(135°\) or \((1,-1)\) direction (downwards and towards 1st point). 3. \(E_3\): At \(45°\) or \((-1,-1)\) direction (upwards and towards 1st point).

Find the Electric Field Components for Each Charge

Let's find the x and y components of the electric fields produced by each charge. We will use trigonometry to do this. $$E_{1x} = E_1 \cos(180°)$$ $$E_{1y} = E_1 \sin(180°)$$ $$E_{2x} = E_2 \cos(135°)$$ $$E_{2y} = E_2 \sin(135°)$$ $$E_{3x} = E_3 \cos(45°)$$ $$E_{3y} = E_3 \sin(45°)$$

Sum the Electric Field Components

Now let's sum the x and y components of the electric field vectors produced by each charge. $$E_{x} = E_{1x} + E_{2x} + E_{3x}$$ $$E_{y} = E_{1y} + E_{2y} + E_{3y}$$

Find the Magnitude of the Total Electric Field

Finally, let's find the magnitude of the total electric field at the point \((3 \mathrm{~m},0)\). We can use the Pythagorean theorem to do this: $$E = \sqrt{E_x^2 + E_y^2}$$ Calculating this quantity will give us the magnitude of the electric field at the given point. Then, compare the result with the given options and select the correct one.

Key concepts

Coulomb's Law

Understanding Coulomb's Law is crucial when dealing with point charges and the resulting electric fields. It quantifies the amount of force between two stationary, electrically charged particles. The law's formula, \[ F = k \frac{|q_1 \cdot q_2|}{r^2} \], where \(F\) is the force in Newtons, \(k\) is Coulomb's constant (\(8.9875 \times 10^9 \ Nm^2C^{-2}\)), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between the charges in meters.

By understanding that the electric field \(E\) is directly related to the force, we can rearrange Coulomb's Law to find the electric field created by a single point charge at a distance \(r\): \[ E = \frac{k \cdot |q|}{r^2} \]. This form highlights that the magnitude of the electric field depends inversely on the square of the distance from the charge and is directly proportional to the charge's magnitude itself.

When utilizing Coulomb's Law for multiple charges, the principle of superposition applies; this allows us to analyze the effect of each charge separately and then sum their contribution to find the total electric field at a given point.

Electric Field Magnitude

The electric field magnitude represents the strength of the electric field at a specific point in space, typically measured in Newtons per Coulomb \(N/C\). The magnitude tells us how strong the electric field is, but it does not give us information about the direction of this field. For a point charge, the magnitude can be calculated using the revised Coulomb's Law formula mentioned earlier.

It is important to note that electric fields exert forces on other charges within the field. The force experienced by a test charge \(q\) in an electric field \(E\) is given by \[ F = qE \. This relationship highlights that the magnitude of the electric field will determine the strength of the force on another charge present within the field.

In the given exercise, after computing the electric fields due to each charge, their individual contributions are vectorially added to get the total electric field at the point of interest. The magnitude of this resultant field is obtained using the Pythagorean theorem to combine the x and y components.

Vector Components

In physics, vector components are an essential part of understanding forces and fields since they provide a means to handle direction and magnitude separately. A vector in a plane can be split into perpendicular components, typically along the x and y axes. The original vector is the sum of these components and can be found using vector addition.

For electric fields, once we have the magnitude, we need to understand the direction to compute the components. The x component \(E_x\) is found by multiplying the magnitude of the field by \(\cos(\theta)\), and the y component \(E_y\) is by multiplying by \(\sin(\theta)\) where \(\theta\) is the angle the original vector makes with the positive x-axis.

In the context of the exercise, the components of each field are calculated considering the direction towards or away from each point charge. Once each component is calculated, they can be summed algebraically to determine the net electric field in each direction. Finally, to determine the total electric field's magnitude, we combine these net components using the Pythagorean theorem.