Electrical Force Between Two Positive Charges

Hey guys! Today, we're diving deep into the fascinating world of physics, specifically tackling a classic problem involving electrostatic forces. You know, the kind of stuff that makes your hair stand on end (sometimes literally!). We've got a scenario with two positive charges, $q_1$ and $q_2$, and we need to figure out the magnitude and direction of the electrical force, $F_e$, that $q_1$ exerts on $q_2$. Let's break it down.

Understanding Coulomb's Law

Before we get our hands dirty with the numbers, let's talk about the fundamental principle at play here: Coulomb's Law. This bad boy governs the interaction between charged particles. It tells us that the electrical 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_e = k rac{|q_1 q_2|}{r^2}

Where:

• $F_e$ is the magnitude of the electrical force.
• $k$ is Coulomb's constant, approximately 8.9875 imes 10^9 N rac{m^2}{C^2}. We often use 9 imes 10^9 N rac{m^2}{C^2} for simplicity in calculations.
• $q_1$ and $q_2$ are the magnitudes of the charges.
• $r$ is the distance between the centers of the two charges.

Now, a crucial point about Coulomb's Law is the direction of the force. Like charges (both positive or both negative) repel each other, meaning the force pushes them apart. Opposite charges (one positive, one negative) attract each other, pulling them closer. In our case, both $q_1$ and $q_2$ are positive, so we're dealing with repulsion. This means $q_1$ will push $q_2$ away, and $q_2$ will push $q_1$ away with an equal and opposite force. This is Newton's third law in action – for every action, there's an equal and opposite reaction!

Applying the Law to Our Problem

Alright, let's get to our specific problem. We are given:

• Charge q_1 = 5 ext{ } oldsymbol{ ext{ extmu}}} C = 5 imes 10^{-6} C (remember, oldsymbol{ ext{ extmu}}} means micro, which is $10^{-6}$).
• Charge q_2 = 2 ext{ } oldsymbol{ ext{ extmu}}} C = 2 imes 10^{-6} C.
• The distance between them, $r = 3 imes 10^{-2} m$.
• $q_1$ is to the west of $q_2$.

We need to find the magnitude and direction of the electrical force, $F_e$, applied by $q_1$ on $q_2$. Since both charges are positive, they repel each other. This means the force exerted by $q_1$ on $q_2$ will be directed away from $q_1$. Since $q_1$ is west of $q_2$, pushing $q_2$ away from $q_1$ means pushing $q_2$ towards the east.

Now, let's calculate the magnitude using Coulomb's Law:

F_e = k rac{|q_1 q_2|}{r^2}

Plugging in our values:

F_e = (9 imes 10^9 N rac{m^2}{C^2}) rac{|(5 imes 10^{-6} C)(2 imes 10^{-6} C)|}{(3 imes 10^{-2} m)^2}

First, let's handle the product of the charges:

$$q_1 q_2 = (5 imes 10^{-6} C)(2 imes 10^{-6} C) = 10 imes 10^{-12} C^2 = 1 imes 10^{-11} C^2$$

Next, let's square the distance:

$$r^2 = (3 imes 10^{-2} m)^2 = 9 imes 10^{-4} m^2$$

Now, substitute these back into the formula:

F_e = (9 imes 10^9 N rac{m^2}{C^2}) rac{1 imes 10^{-11} C^2}{9 imes 10^{-4} m^2}

We can simplify this:

F_e = rac{9 imes 10^9 imes 1 imes 10^{-11}}{9 imes 10^{-4}} N

F_e = rac{9 imes 10^{-2}}{9 imes 10^{-4}} N

$$F_e = 1 imes 10^{-2 - (-4)} N$$

$$F_e = 1 imes 10^{2} N$$

$$F_e = 100 N$$

So, the magnitude of the electrical force applied by $q_1$ on $q_2$ is 100 Newtons. Remember, the direction is towards the east because like charges repel, and $q_1$ is west of $q_2$. Pretty neat, huh?

Factors Affecting Electrical Force

It's super important to remember that the electrical force isn't some mystical constant; it's influenced by a few key factors, all encapsulated in Coulomb's Law. Let's dive into what makes this force tick. The magnitude of the charges is a big one, guys. The formula shows that the force is directly proportional to the product of the charges ($q_1 q_2$). This means if you double the size of either charge, the force between them doubles. If you double both charges, the force quadruples! It's a pretty sensitive relationship. Imagine two magnets; the stronger the magnets, the stronger the pull or push. It's a similar concept with electric charges. When dealing with oldsymbol{ ext{microcoulombs}} (oldsymbol{ ext{ extmu}}}C), even small-seeming differences in charge can lead to noticeable forces, especially when other factors are in play. Our $q_1$ is 5 oldsymbol{ ext{ extmu}}}C and $q_2$ is 2 oldsymbol{ ext{ extmu}}}C. If $q_1$ were, say, 10 oldsymbol{ ext{ extmu}}}C (double its current value) and $q_2$ remained 2 oldsymbol{ ext{ extmu}}}C, the force would double. If $q_1$ were 5 oldsymbol{ ext{ extmu}}}C and $q_2$ were 4 oldsymbol{ ext{ extmu}}}C, the force would again double. If both were doubled, the force would quadruple. It’s crucial to keep these charge values accurate for precise calculations.

Then there's the distance between the charges, represented by $r$. This is where things get a bit more dramatic. The force is inversely proportional to the square of the distance ($r^2$). This means if you double the distance between two charges, the force between them doesn't just get halved; it gets reduced to one-fourth of its original value ($1/(2^2) = 1/4$). Conversely, if you halve the distance, the force increases by a factor of four ($1/(1/2)^2 = 1/(1/4) = 4$). This inverse square relationship is a hallmark of many fundamental forces in nature, including gravity. It implies that electrical forces weaken rapidly as charges move farther apart. In our problem, the distance is $3 imes 10^{-2} m$, or 3 centimeters. If we moved these charges to 6 centimeters apart (doubling the distance), the force would drop from 100 N to about 25 N. If we brought them closer, say to 1.5 centimeters (halving the distance), the force would skyrocket to 400 N! This sensitivity to distance is why phenomena like static electricity are more noticeable when objects are close together.

Finally, we have Coulomb's constant ($k$). This constant is a fundamental property of the universe, reflecting the strength of the electromagnetic force in a vacuum. Its value is approximately 8.9875 imes 10^9 N rac{m^2}{C^2}. We often approximate it as 9 imes 10^9 N rac{m^2}{C^2} for easier calculations, as we did. This constant ensures that the units work out correctly and provides the correct proportionality. Think of it as a scaling factor that dictates how strong the electrical force is in our everyday environment (or in a vacuum, to be precise). It's a universal constant, meaning it's the same everywhere in the universe, unlike the charges or distances which can vary wildly.

The Role of Medium

It’s also worth noting that Coulomb's Law, in the form we've used, applies to charges in a vacuum. If the charges are placed in a different medium, like water or oil, the force between them can be reduced. This is because the medium can become polarized, with its own charges rearranging to partially shield the original charges from each other. This effect is quantified by the dielectric constant of the medium. The force in a medium is given by F_e = rac{1}{ ext{dielectric constant}} k rac{|q_1 q_2|}{r^2}. So, the force is weaker in materials other than a vacuum. For instance, the dielectric constant of water is about 80, meaning the force between charges in water would be about 80 times weaker than in a vacuum, assuming the same charges and distance.

Direction Matters: Repulsion vs. Attraction

Guys, let's hammer home the point about direction. This is where a lot of people get tripped up. In our specific problem, we have two positive charges. Remember the golden rule: like charges repel, opposite charges attract. Since $q_1$ and $q_2$ are both positive, they push each other away. The question asks for the force applied by $q_1$ on $q_2$. We know $q_1$ is located to the west of $q_2$. For $q_1$ to push $q_2$ away from itself, the force must be directed eastward. Imagine you're standing between two strong magnets with their north poles facing each other; you'd feel them pushing you apart. If one magnet is to your left and the other to your right, the left magnet pushes you right, and the right magnet pushes you left. In our case, $q_1$ is the