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A particle has charge \(-5.00 \mathrm{nC}\). (a) Find the magnitude and direction of the electric field due to this particle at a point \(0.250 \mathrm{~m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of \(12.0 \mathrm{~N} / \mathrm{C} ?\)

Short Answer

Expert verified
The magnitude and the direction of the electric field due to the particle at a point 0.250 m directly above it are \( 7.19×10^4 N/C \) and towards the particle, respectively. The distance from the particle where its electric field has a magnitude of \( 12.0 N/C \) is about 0.0654 m.

Step by step solution

01

Find the magnitude of the electric field at a specific distance

\(E = \frac{k_e|q|}{r^2}\), where \( k_e \) is Coulomb's constant (\( 8.99×10^9 N m²/C² \)), |q| is the absolute value of the charge (| - 5.00 nC | = \( 5.00×10^{-9} C \)), and \( r \) is the distance (0.250 m). Now, just plug these values into the formula to find the electric field.
02

Find the direction of the electric field at a specific distance

The electric field direction is always from positive charge to negative charge. Here the charge is negative, so the electric field direction is towards the particle.
03

Find the distance from the particle at a given electric field

We can rearrange the formula we used in step 1 to \( r = \sqrt{\frac{k_e|q|}{E}} \), Now plug the given magnitude of the electric field (\( 12.0 N/C \)) and the absolute value of the charge into this formula to find the distance r,

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Constant
Coulomb's constant, often denoted as \( k_e \), is a fundamental value in physics used to quantify the strength of the electric force between charged particles. It appears in Coulomb's Law, which calculates the electric force between two point charges. The constant is defined as \( k_e = 8.99\times10^9 \, \text{N m}^2/\text{C}^2 \).
This constant is crucial when determining the electric field due to a charge at a given point in space. In calculations, the electric field \( E \) is given by the formula \( E = \frac{k_e |q|}{r^2} \), where \( |q| \) is the magnitude of a charge and \( r \) is the distance from the charge. Knowing Coulomb's constant allows us to compute the electric field's magnitude, which is a measure of how strong the field is at a certain point relative to the charged object.
In the context of the example problem, using \( k_e \) ensures precision when evaluating the electric interactions at play, making it possible to predict how charged particles will influence their surroundings.
Electric Charge
Electric charge is a fundamental property of matter, which causes it to experience a force when placed in an electric or magnetic field. Charge is quantized and comes in two types: positive and negative. In calculations, charge is often expressed in Coulombs (C). It determines not just the amount but also the direction of the electric field emanating from it.
In the exercise, the particle has a charge of \(-5.00 \, \text{nC}\). The negative sign indicates that the charge is negative, meaning the electric field lines will point towards the charge. By understanding the properties of the particle's electric charge, we can predict the electric field's direction. If the charge had been positive, the field lines would point away from it.
The electric field around a point charge is represented by \( E = \frac{k_e |q|}{r^2} \). Here, \( |q| \) is the absolute value of the charge, which excludes the negative sign, showing the magnitude of the charge without its direction.
Distance in Electric Fields
Distance plays a crucial role in determining the magnitude of the electric field around a charged particle. In the formula \( E = \frac{k_e |q|}{r^2} \), the variable \( r \) represents this distance. Importantly, distance is inversely proportional to the square of the field's magnitude. This means the farther you are from the charge, the weaker the electric field becomes.
Understanding the relationship between distance and electric fields helps us solve problems like part (b) of our exercise, where we determine how far from the particle the field has a specific strength, such as \( 12.0 \, \text{N/C} \).
To find this distance, we rearrange the formula to \( r = \sqrt{\frac{k_e |q|}{E}} \). Solving this allows us to explore the spatial dimensions of electric fields, effectively calculating how influence diminishes with increasing separation from the charge.

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Most popular questions from this chapter

If two electrons are each \(1.50 \times 10^{-10} \mathrm{~m}\) from a proton (Fig. E21.41), find the magnitude and direction of the net electric force they will exert on the proton.

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A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x=0, y=-1.35 \mathrm{~m}\) (b) at \(x=12.0 \mathrm{~cm}, y=12.0 \mathrm{~cm} ;\) (c) at \(x=-1.10 \mathrm{~m}, y=2.60 \mathrm{~m} ?\) Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\)

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