Electric Forces: Finding Resultant Force On Q0

Hey guys! Today, we're diving into an exciting problem involving electric forces. We'll be looking at how to figure out the forces acting on a charge due to other charges around it. Specifically, we'll tackle a scenario where we have a charge $q_0$ sitting at a point C, and we want to find out what happens when two other charges, $q_1$ and $q_2$, are placed nearby at points A and B, respectively. This is a classic physics problem that helps us understand Coulomb's Law and the principle of superposition of forces. So, let's jump right in!

Understanding the Problem

Before we start crunching numbers, let's make sure we understand the situation. We have three charges:

• $q_0 = +0.4 nC$ (nanoCoulombs) located at point C.
• $q_1 = -5 nC$ located at point A.
• $q_2 = +5 nC$ located at point B.

The distances are:

• $AC = 3 cm$
• $CB = 3 cm$

The goal is to find the direction and magnitude of the net (resultant) force acting on $q_0$ due to the other two charges, $q_1$ and $q_2$. To do this, we'll need to use Coulomb's Law, which describes the force between two point charges.

Coulomb's Law: A Quick Recap

Coulomb's Law is the cornerstone of electrostatics, stating that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it's expressed as:

$F = k * |q_1 * q_2| / r^2$

Where:

• $F$ is the magnitude of the electric force.
• $k$ is Coulomb's constant, approximately $8.9875 × 10^9 N⋅m^2/C^2$.
• $q_1$ and $q_2$ are the magnitudes of the charges.
• $r$ is the distance between the charges.

It's crucial to remember that force is a vector quantity, meaning it has both magnitude and direction. The direction of the force is along the line joining the two charges. If the charges have the same sign (both positive or both negative), the force is repulsive. If the charges have opposite signs, the force is attractive. This understanding of attractive and repulsive forces is essential for determining the direction of the forces acting on our charge $q_0$.

Step 1: Finding the Forces due to Individual Charges

Our first step is to calculate the forces acting on $q_0$ due to each of the other charges individually. We'll calculate the force exerted by $q_1$ on $q_0$ ($F_{10}$) and the force exerted by $q_2$ on $q_0$ ($F_{20}$).

Force due to $q_1$ ($F_{10}$)

$q_1$ has a negative charge (-5 nC) and $q_0$ has a positive charge (+0.4 nC). Opposite charges attract, so $F_{10}$ will be an attractive force, pulling $q_0$ towards $q_1$. Since $q_1$ is located at point A, this force will be directed along the line AC.

Let's calculate the magnitude of $F_{10}$ using Coulomb's Law:

$F_{10} = k * |q_1 * q_0| / AC^2$

First, we need to convert the charges to Coulombs and the distance to meters:

• $q_1 = -5 nC = -5 × 10^{-9} C$
• $q_0 = +0.4 nC = +0.4 × 10^{-9} C$
• $AC = 3 cm = 0.03 m$

Now, plug in the values:

$F_{10} = (8.9875 × 10^9 N⋅m^2/C^2) * |(-5 × 10^{-9} C) * (0.4 × 10^{-9} C)| / (0.03 m)^2$ $F_{10} ≈ (8.9875 × 10^9) * (2 × 10^{-18}) / (0.0009)$ $F_{10} ≈ 1.997 × 10^{-5} N$

So, the magnitude of the force $F_{10}$ is approximately $1.997 × 10^{-5} N$, and its direction is along the line AC, pulling $q_0$ towards A.

Force due to $q_2$ ($F_{20}$)

$q_2$ has a positive charge (+5 nC) and $q_0$ also has a positive charge (+0.4 nC). Like charges repel, so $F_{20}$ will be a repulsive force, pushing $q_0$ away from $q_2$. Since $q_2$ is located at point B, this force will be directed along the line CB, away from B.

Let's calculate the magnitude of $F_{20}$ using Coulomb's Law:

$F_{20} = k * |q_2 * q_0| / CB^2$

We already have the values in Coulombs and meters:

• $q_2 = +5 × 10^{-9} C$
• $q_0 = +0.4 × 10^{-9} C$
• $CB = 0.03 m$

Now, plug in the values:

$F_{20} = (8.9875 × 10^9 N⋅m^2/C^2) * |(5 × 10^{-9} C) * (0.4 × 10^{-9} C)| / (0.03 m)^2$ $F_{20} ≈ (8.9875 × 10^9) * (2 × 10^{-18}) / (0.0009)$ $F_{20} ≈ 1.997 × 10^{-5} N$

So, the magnitude of the force $F_{20}$ is also approximately $1.997 × 10^{-5} N$, and its direction is along the line CB, pushing $q_0$ away from B.

Step 2: Finding the Resultant Force

Now that we have the individual forces $F_{10}$ and $F_{20}$, we need to find the resultant force acting on $q_0$. Since forces are vectors, we need to consider both their magnitudes and directions. The resultant force is the vector sum of all the individual forces.

In our case, $F_{10}$ pulls $q_0$ towards A, and $F_{20}$ pushes $q_0$ away from B. We found that the magnitudes of $F_{10}$ and $F_{20}$ are equal ($1.997 × 10^{-5} N$). This simplifies our calculation significantly!

Visualizing the Forces

Imagine points A, B, and C forming a triangle. We know that AC = CB = 3 cm, which means triangle ABC is at least an isosceles triangle. To find the resultant force, we need to know the angle between the forces $F_{10}$ and $F_{20}$. Without additional information about the geometry (like the angle ACB), we can't determine the exact angle between the forces and, therefore, can't find the precise magnitude of the resultant force using vector addition components (like cosine rule). However, if we assume that A, C, and B lie on a straight line (which is a common scenario in these types of problems), the forces will be acting along the same line, making the calculation much simpler.

Assuming A, C, and B are Collinear

Let's assume that points A, C, and B are arranged in a straight line, with C between A and B. In this case, the forces $F_{10}$ and $F_{20}$ act along the same line, but in opposite directions. Since their magnitudes are equal, the resultant force will be the difference between their magnitudes:

$F_{resultant} = |F_{20} - F_{10}|$

Since $F_{10} = F_{20} = 1.997 × 10^{-5} N$,

$F_{resultant} = |1.997 × 10^{-5} N - 1.997 × 10^{-5} N| = 0 N$

Therefore, if A, C, and B are collinear, the resultant force on $q_0$ is 0 N.

General Case: Vector Addition

If A, C, and B are not collinear, we need to use vector addition to find the resultant force. This involves breaking down the forces into their components (x and y) and then adding the components separately. However, without knowing the angle ACB, we cannot proceed with this method. Let's assume for the sake of demonstration that angle ACB is θ. Then, we can use the parallelogram method or the component method for vector addition.

1. Parallelogram Method: Construct a parallelogram with $F_{10}$ and $F_{20}$ as adjacent sides. The diagonal of the parallelogram starting from point C represents the resultant force in both magnitude and direction.
2. Component Method:
   • Resolve $F_{10}$ and $F_{20}$ into their x and y components.
   • Add the x-components to get the x-component of the resultant force ($F_{x}$).
   • Add the y-components to get the y-component of the resultant force ($F_{y}$).
   • The magnitude of the resultant force is $F_{resultant} = \sqrt{F_x^2 + F_y^2}$.
   • The direction of the resultant force can be found using $\tan^{-1}(F_y / F_x)$.

If we assume angle ACB is θ, and we denote the forces $F_{10}$ and $F_{20}$ as vectors, we can add them:

$\vec{F}_{resultant} = \vec{F}_{10} + \vec{F}_{20}$

Let's express the forces in component form. If we consider the line CB as the x-axis and a line perpendicular to it at C as the y-axis, then:

$\vec{F}_{20} = F_{20} \hat{i} = 1.997 × 10^{-5} \hat{i} N$ $\vec{F}_{10} = F_{10} \cos(θ) \hat{i} + F_{10} \sin(θ) \hat{j} = -1.997 × 10^{-5} \cos(θ) \hat{i} + 1.997 × 10^{-5} \sin(θ) \hat{j} N$

The negative sign for the x-component of $F_{10}$ is because it acts in the opposite direction to $F_{20}$. Now we can add the vectors:

$\vec{F}_{resultant} = (1.997 × 10^{-5} - 1.997 × 10^{-5} \cos(θ)) \hat{i} + (1.997 × 10^{-5} \sin(θ)) \hat{j} N$

The magnitude of the resultant force is:

$|F_{resultant}| = \sqrt{((1.997 × 10^{-5} - 1.997 × 10^{-5} \cos(θ))^2 + (1.997 × 10^{-5} \sin(θ))^2)}$

Example: If θ = 90°

If the angle ACB is 90 degrees, then:

$|F_{resultant}| = \sqrt{((1.997 × 10^{-5})^2 + (1.997 × 10^{-5})^2)} ≈ 2.824 × 10^{-5} N$

And the direction can be calculated using the arctangent of the components.

Conclusion

So, to recap, we found the forces acting on charge $q_0$ due to charges $q_1$ and $q_2$ using Coulomb's Law. We calculated the individual forces and then considered their vector sum to find the resultant force. If the charges are collinear, and equal magnitude attractive and repulsive forces cancel each other out resulting in a zero net force. If not collinear, vector addition (either by components or parallelogram method) is needed, which requires knowing the angle between the forces.

This problem illustrates the fundamental principles of electrostatics and the importance of understanding vector addition when dealing with forces. I hope this explanation was helpful, guys! If you have any more questions, feel free to ask! Remember, the key is to break down the problem into smaller, manageable steps and apply the fundamental laws of physics. Keep practicing, and you'll become a pro at solving these types of problems in no time!