Problem 24 - College Physics

Question

What are the magnitude and direction of the electric field at a point \(0.75 \mathrm{~cm}\) away from a point charge of \(+2.0 \mathrm{pC} ?\)

Step-by-Step Solution

Answer

The electric field magnitude is approximately 319.64 N/C and it points away from the charge.

Step 1: Understand the Problem

We need to find the magnitude and direction of the electric field created by a point charge at a specific distance. A point charge of +2.0 pC (picoCoulombs) is given, and the point is 0.75 cm away from the charge.

Step 2: Recall the Electric Field Formula

The magnitude of the electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula: \[ E = \frac{k \cdot |q|}{r^2} \]where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).

Step 3: Convert Units

Convert the distance from cm to meters, since SI units are needed for the calculation: \[ 0.75 \text{ cm} = 0.0075 \text{ m} \] Also convert the charge from picoCoulombs to Coulombs: \[ q = 2.0 \text{ pC} = 2.0 \times 10^{-12} \text{ C} \]

Step 4: Plug Values into the Formula

Substitute the known values into the electric field formula: \[ E = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 2.0 \times 10^{-12} \text{ C}}{(0.0075 \text{ m})^2} \] Calculate the result to find the magnitude.

Step 5: Calculate the Magnitude

Compute the electric field magnitude: \[ E = \frac{8.99 \times 10^9 \times 2.0 \times 10^{-12}}{0.0075^2} = \frac{1.798 \times 10^{-2}}{0.00005625} \approx 319.64 \text{ N/C} \]

Step 6: Determine the Direction

Since the charge is positive, the electric field direction is away from the charge. Therefore, at the point 0.75 cm away, the electric field points radially outward.

Key Concepts

Coulomb's Law

Coulomb's law helps us understand the interaction between charged particles. It describes how the electric force between two charges relates to their distance. Imagine you have two tiny charged particles, called point charges. The force is strong when the charges are close, but it weakens as they move apart. The formula is:

• Force, \( F = k \frac{|q_1 q_2|}{r^2} \)

where \( k \) is a very large constant, about \( 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \), helping express the strong force between charges.
The charges \( q_1 \) and \( q_2 \) are the magnitudes of the point charges, and \( r \) is the separation distance.
Since the force relates directly to the electric field, understanding Coulomb's law is key to calculating the field around a point charge.

Point Charge

A point charge is a model used in physics to describe a charged object by assuming all its charge is concentrated at a single point. This simplifies calculations and helps us predict electric fields without needing the shape or size of the object.
For instance, if you have a tiny particle with a charge of \(+2.0 \text{ pC}\), we can consider it a point charge for calculating the electric field.
This concept is crucial because it allows us to use simple equations to model the electric properties around the charge. We use the point charge model in situations where the distances involved are large compared to the size of the charged object.

Unit Conversion

Unit conversion is an essential skill in physics. It's about translating measurements from one unit to another. The electric field formula needs SI units, so we must convert the given values accordingly.
Distances are usually best expressed in meters, and charges in Coulombs. In the exercise, \(0.75 \text{ cm}\) was converted to \(0.0075 \text{ m}\), and \(2.0 \text{ pC}\) was converted to \(2.0 \times 10^{-12} \text{ C}\).
Converting units might seem tedious but helps ensure calculations are accurate, allowing us to apply formulas like the electric field formula effectively.