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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

A semicircle of radius \(a\) is in the first and second quadrants, with the center of curvature at the origin. Positive charge \(+Q\) is distributed uniformly around the left half of the semicircle, and negative charge \(-Q\) is distributed uniformly around the right half of the semicircle (Fig. P21.84). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?

Two small spheres spaced \(20.0 \mathrm{~cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(3.33 \times 10^{-21} \mathrm{~N} ?\)

Two thin rods, each with length \(L\) and total charge \(+Q,\) are parallel and separated by a distance \(a .\) The first rod has one end at the origin and its other end on the positive \(y\) -axis. The second rod has its lower end on the positive \(x\) -axis. (a) Explain why the \(y\) -component of the net force on the second rod vanishes. (b) Determine the \(x\) -component of the differential force \(d F_{2}\) exerted on a small portion of the second rod, with length \(d y_{2}\) and position \(y_{2},\) by the first rod. (This requires integrating over differential portions of the first rod, parameterized by \(\left.d y_{1} .\right)\) (c) Determine the net force \(\vec{F}_{2}\) on the second rod by integrating \(d F_{2 x}\) over the second rod. (d) Show that in the limit \(a \gg L\) the force determined in part (c) becomes \(\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{a^{2}} \hat{\imath}\). (e) Determine the external work required to move the second rod from very far away to the position \(x=a\), provided the first rod is held fixed at \(x=0 .\) This describes the potential energy of the original configuration. (f) Suppose \(L=50.0 \mathrm{~cm}, a=10.0 \mathrm{~cm}, Q=10.0 \mu \mathrm{C},\) and \(m=500 \mathrm{~g}\). If the two rods are released from the original configuration, they will fly apart and ultimately achieve a particular relative speed. What is that relative speed?

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{C}\) is placed at the point \(x=+0.150 \mathrm{~m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{~m}, y=0 .\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{~m}, y=0\) (c) \(x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m}\) (d) \(x=0, y=0.200 \mathrm{~m}\)

An American penny is \(97.5 \%\) zinc and \(2.5 \%\) copper and has a mass of \(2.5 \mathrm{~g}\). (a) Use the approximation that a penny is pure zinc, which has an atomic mass of \(65.38 \mathrm{~g} / \mathrm{mol},\) to estimate the number of electrons in a penny. (Each zinc atom has 30 electrons.) (b) Estimate the net charge on all of the electrons in one penny. (c) The net positive charge on all of the protons in a penny has the same magnitude as the charge on the electrons. Estimate the force on either of two objects with this net magnitude of charge if the objects are separated by \(2 \mathrm{~cm}\). (d) Estimate the number of leaves on an oak tree that is 60 feet tall. (e) Imagine a forest filled with such trees, arranged in a square lattice, each \(10 \mathrm{~m}\) distant from its neighbors. Estimate how large such a forest would need to be to include as many leaves as there are electrons in one penny. (f) How does that area compare to the surface area of the earth?

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