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Chapter 22: Problem 25

\( \mathrm{~A}+48.00-\mathrm{nC}\) point charge is placed on the \(x\) -axis at \(x=4.000 \mathrm{~m},\) and \(\mathrm{a}-24.00-\mathrm{n} \mathrm{C}\) point charge is placed on the \(y\) -axis at \(y=-6.000 \mathrm{~m} .\) What is the direction of the electric field at the origin?

Short Answer

Expert verified
Answer: The direction of the net electric field at the origin can be found using the formula \(\theta = \arctan(\frac{E_{nety}}{E_{netx}}) = \arctan(\frac{E_B}{-E_A})\), where \(E_A\) and \(E_B\) are the electric fields due to charges A and B, respectively. Plug in the values of A and B, calculate \(E_A\) and \(E_B\), and then find the angle \(\theta\) to determine the direction of the net electric field at the origin.

Step by step solution

01

Finding the electric field due to charge A at the origin

To find the electric field due to the first charge A at the origin, use the formula for the electric field from a point charge: \(E = k* \frac{Q}{r^2}\), where \(k = 8.9875517923*10^9 N m^2/C^2\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance between the charge and the point at which we want to find the electric field. The distance between charge A and the origin is \(x=4m\). Therefore, the electric field at the origin due to charge A is: \(E_A = k* \frac{A}{(4m)^2}\).
02

Finding the electric field due to charge B at the origin

Repeat the process for charge B, which is placed on the y-axis at \(y=-6m\). The distance between charge B and the origin is \(6m\). Therefore, the electric field at the origin due to charge B is: \(E_B = k* \frac{B}{(6m)^2}\), where B is -24.00 nC.
03

Determining the components of the electric fields

We know that the electric field due to charge A is in the negative x-direction and the electric field due to charge B is in the positive y-direction. Thus, the components of the electric fields are: \(E_{Ax}=-E_A, E_{Ay}=0\) \(E_{Bx}=0, E_{By}=E_B\)
04

Vector addition of the electric fields

Now, we'll find the net electric field at the origin by adding the components of the electric fields due to charges A and B: \(E_{netx} = E_{Ax} + E_{Bx} = -E_A\) \(E_{nety} = E_{Ay} + E_{By} = E_B\)
05

Calculating the direction of the net electric field

To find the direction of the net electric field at the origin, we need to determine the angle with respect to the x-axis. We use the arctangent function to find the angle: \(\theta = \arctan(\frac{E_{nety}}{E_{netx}}) = \arctan(\frac{E_B}{-E_A})\) Now, you can plug in the values of A, B and find \(E_A\) and \(E_B\) to calculate the angle, \(\theta\), which gives the direction of the net electric field at the origin.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics that describes the interaction between two point charges. It states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is expressed as:\[F = k \frac{|Q_1Q_2|}{r^2}\] where:
  • \(F\) is the magnitude of the force between the charges,
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \mathrm{N\cdot m^2/C^2}\),
  • \(Q_1\) and \(Q_2\) are the amounts of the two charges, and
  • \(r\) is the distance between the centers of the two charges.
Using this law, one can not only calculate the force but also derive the electric field created by a point charge. For instance, replacing the force \(F\) with the electric field \(E\), we use the equation:\[E = k \frac{Q}{r^2}\]This formula helps in predicting electric fields generated by charges alone.
Vector Addition
Vector addition is essential when dealing with electric fields, especially when multiple charges are present. Since electric fields are vector quantities, they have both a magnitude and direction.
When you are trying to find the net electric field due to multiple charges, each of the electric fields must be treated as vectors.
Let's break it down:
  • Determine the magnitude and direction of each electric field component based on its charge鈥檚 sign and position.
  • Use trigonometry to resolve electric fields into their x and y components, if necessary.
  • Add the components in each direction separately: sum up all x-components for a total x-component, do the same with y components.
The result is a combined vector.
Finally, to get the overall direction, you can calculate the angle using the arctangent function: \[\theta = \arctan \left( \frac{E_{net_y}}{E_{net_x}} \right)\]This helps in understanding how different fields interact at a point.
Point Charge Electric Field
A point charge creates an electric field that radiates outwards (if positive) or inwards (if negative).
This field embodies how the charge influences other charges around it.
The electric field at any point in space produced by a single point charge can be determined by the formula:\[E = k \frac{Q}{r^2}\]The direction of the field is radially away from the charge if the charge is positive, and towards the charge if it is negative.
  • This radial pattern implies that the field decreases in strength as the square of the distance from the charge, meaning it weakens rapidly with increasing distance.
  • Electric fields are especially important at the origin (reference point) when taking contribution from multiple point charges.
  • Understanding these fields helps visualize how an arrangement of point charges affects a point at any location in space.
Point charge electric fields lay the foundation for learning more advanced field interactions and configurations, like dipoles or continuous charge distributions.

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Most popular questions from this chapter

A solid nonconducting sphere of radius \(a\) has a total charge \(+Q\) uniformly distributed throughout its volume. The surface of the sphere is coated with a very thin (negligible thickness) conducting layer of gold. A total charge of \(-2 Q\) is placed on this conducting layer. Use Gauss's Law to do the following. a) Find the electric field \(E(r)\) for \(ra\) (outside the coated sphere, beyond the sphere and the gold layer). c) Sketch the graph of \(E(r)\) versus \(r\). Comment on the continuity or discontinuity of the electric field, and relate this to the surface charge distribution on the gold layer.

A hollow conducting spherical shell has an inner radius of \(8.00 \mathrm{~cm}\) and an outer radius of \(10.0 \mathrm{~cm} .\) The electric field at the inner surface of the shell, \(E_{\mathrm{i}}\), has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points toward the center of the sphere, and the electric field at the outer surface, \(E_{\infty}\) has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points away from the center of the sphere (see the figure). Determine the magnitude of the charge on the inner surface and the outer surface of the spherical shell.

A solid nonconducting sphere has a volume charge distribution given by \(\rho(r)=(\beta / r) \sin (\pi r / 2 R) .\) Find the total charge contained in the spherical volume and the electric field in the regions \(rR\). Show that the two expressions for the electric field equal each other at \(r=R\).

A single positive point charge, \(q,\) is at one corner of a cube with sides of length \(L\), as shown in the figure. The net electric flux through the three net electric flux through the three adjacent sides is zero. The net electric flux through each of the other three sides is a) \(q / 3 \epsilon_{0}\). b) \(q / 6 \epsilon_{0}\). c) \(q / 24 \epsilon_{0}\). d) \(q / 8 \epsilon_{0}\).

An infinitely long, solid cylinder of radius \(R=9.00 \mathrm{~cm},\) with a uniform charge per unit of volume of \(\rho=6.40 \cdot 10^{-8} \mathrm{C} / \mathrm{m}^{3},\) is centered about the \(y\) -axis. Find the magnitude of the electric field at a radius \(r=4.00 \mathrm{~cm}\) from the center of this cylinder.
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