Electric Field Intensity: 16μC Charge At 4 Meters

Hey guys! Ever wondered how strong the electric field is around a charged object? Today, we're diving deep into calculating the electric field intensity created by a point charge. Specifically, we'll tackle a scenario involving a 16 microcoulomb (16μC) charge and a point 4 meters away. Let's break it down step-by-step, making sure we understand the physics behind it and how to apply the relevant formulas. Buckle up, because we're about to get electrified (pun intended!).

Understanding Electric Fields

First things first, what exactly is an electric field? Imagine a charged object sitting in space. It creates an invisible field around itself, a region where other charged objects would experience a force. This force can be either attractive (if the charges are opposite) or repulsive (if the charges are the same). The strength of this field at a particular point is what we call the electric field intensity, often denoted by the symbol 'E'.

The electric field is a vector quantity, meaning it has both magnitude (strength) and direction. The direction of the electric field at a point is defined as the direction of the force that a positive test charge would experience if placed at that point. So, for a positive point charge, the electric field lines radiate outwards, while for a negative point charge, they point inwards. This is a crucial concept to visualize when dealing with electric fields.

Now, to really grasp this, think about magnets. They have magnetic fields around them, and you can feel the force when you bring another magnet close. Electric fields are similar, but instead of magnets, we're dealing with electric charges. The bigger the charge, the stronger the electric field it creates. And the closer you are to the charge, the stronger the field you'll experience. This intuitive understanding will help us as we move on to the math.

The Formula for Electric Field Intensity

The key to calculating electric field intensity is the formula derived from Coulomb's Law. Coulomb's Law, in essence, tells us the force between two point charges. From that, we can derive the electric field intensity (E) due to a single point charge (Q) at a distance (r) away. The formula is:

E = k * |Q| / r²

Where:

• E is the electric field intensity (measured in Newtons per Coulomb, N/C)
• k is Coulomb's constant (approximately 8.99 × 10^9 N⋅m²/C²)
• |Q| is the absolute value of the charge (measured in Coulombs, C)
• r is the distance from the charge to the point where we're calculating the field (measured in meters, m)

Let's break this down piece by piece. The 'k' is a constant that ensures our units work out correctly. The '|Q|' signifies that we only care about the magnitude of the charge, not its sign (whether it's positive or negative) for calculating the field strength. The 'r²' in the denominator is super important – it tells us that the electric field intensity decreases rapidly as we move further away from the charge (inversely proportional to the square of the distance!). So, double the distance, and the field strength drops to a quarter of its original value. This inverse square relationship is fundamental in physics and shows up in many different contexts.

Applying the Formula to Our Problem

Alright, with the theory and the formula under our belts, let's get practical. Our problem states that we have a 16μC (microcoulomb) point charge and we want to find the electric field intensity at a point A, which is 4 meters away.

First, we need to convert microcoulombs (μC) to Coulombs (C). Remember that 'micro' means one millionth, so 1 μC = 1 × 10^-6 C. Therefore, our charge Q is:

Q = 16 μC = 16 × 10^-6 C

Now we have all the pieces we need. We know:

• k = 8.99 × 10^9 N⋅m²/C²
• |Q| = 16 × 10^-6 C
• r = 4 m

Let's plug these values into our formula:

E = (8.99 × 10^9 N⋅m²/C²) * (16 × 10^-6 C) / (4 m)²

Time to whip out the calculator (or do some careful mental math!). Let's simplify step by step:

E = (8.99 × 10^9) * (16 × 10^-6) / 16 N/C

Notice that the 16 in the numerator and denominator conveniently cancel out:

E = 8.99 × 10^9 × 10^-6 N/C

Now we just need to combine the powers of 10:

E = 8.99 × 10^(9-6) N/C

E = 8.99 × 10^3 N/C

So, the electric field intensity at point A is approximately 8.99 × 10^3 N/C, or 8990 N/C. That's a pretty strong electric field! This result tells us the force that a 1 Coulomb positive charge would experience if placed at point A. The direction of this force would be radially outwards from the 16μC charge, since it's a positive charge.

Key Takeaways and Implications

Let's recap what we've learned. We started by understanding the concept of an electric field and electric field intensity. We then introduced the formula for calculating the electric field intensity due to a point charge: E = k * |Q| / r². We carefully identified each variable, converted units when necessary, and plugged the values into the formula. We then performed the calculations step by step to arrive at our answer: 8990 N/C.

This example highlights a few key takeaways:

• The electric field intensity is directly proportional to the charge. Larger charges create stronger electric fields.
• The electric field intensity is inversely proportional to the square of the distance. The further away you are from the charge, the weaker the field.
• Understanding units and conversions is crucial in physics problems. Always make sure your units are consistent before plugging them into a formula.
• The electric field is a vector quantity, so it has both magnitude and direction. While we calculated the magnitude here, remembering the direction (radially outwards for a positive charge) is important for a complete understanding.

Real-World Applications

You might be wondering,