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Chapter 16: Problem 63

In a uniform electric field of magnitude \(E\), the field lines cross through a rectangle of area \(A\) at an angle of \(60.0^{\circ}\) with respect to the plane of the rectangle. What is the flux through the rectangle?

Short Answer

Expert verified
Answer: The electric flux through the rectangle is \(\frac{1}{2} \times E \times A\).

Step by step solution

01

Identify the given values and formula to be used

We are provided with the following information: - Magnitude of the uniform electric field, \(E\). - Area of the rectangle, \(A\). - Angle (relative to the plane of the rectangle), \(\theta = 60.0^{\circ}\). The formula needed to calculate the electric flux is: \(\Phi = EA \cos{\theta}\).
02

Convert the angle from degrees to radians

Before using the formula, we need to convert the angle \(\theta\) from degrees to radians: \(\theta_\text{radians} = \theta_\text{degrees} \times \frac{\pi}{180}\)
03

Calculate the angle in radians

Using the given angle of \(60.0^{\circ}\), we can now find the angle in radians: \(\theta_\text{radians} = 60.0 \times \frac{\pi}{180} = \frac{\pi}{3}\)
04

Apply the formula to find electric flux

Now let's plug the values into the flux formula, \(\Phi = EA \cos{\theta_\text{radians}}\): \(\Phi = E \times A \times \cos{(\frac{\pi}{3})}\)
05

Solve for the electric flux

Evaluate the expression for the electric flux: \(\Phi = EA (\frac{1}{2})\) \(\Phi = \frac{1}{2} \times E \times A\) The electric flux through the rectangle is \(\frac{1}{2} \times E \times A\).

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Most popular questions from this chapter

How many electrons must be removed from each of two 5.0 -kg copper spheres to make the electric force of repulsion between them equal in magnitude to the gravitational attraction between them?
A charge of \(63.0 \mathrm{nC}\) is located at a distance of \(3.40 \mathrm{cm}\) from a charge of \(-47.0 \mathrm{nC} .\) What are the \(x\) - and \(y\) -components of the electric field at a point \(P\) that is directly above the 63.0 -nC charge at a distance of \(1.40 \mathrm{cm} ?\) Point \(P\) and the two charges are on the vertices of a right triangle.
Using the results of Problem \(66,\) we can find the electric field at any radius for any spherically symmetrical charge distribution. A solid sphere of charge of radius \(R\) has a total charge of \(q\) uniformly spread throughout the sphere. (a) Find the magnitude of the electric field for \(r \geq R .\) (b) Find the magnitude of the electric field for \(r \leq R .\) (c) Sketch a graph of \(E(r)\) for \(0 \leq r \leq 3 R\)
A small sphere with a charge of \(-0.60 \mu \mathrm{C}\) is placed in a uniform electric field of magnitude \(1.2 \times 10^{6} \mathrm{N} / \mathrm{C}\) pointing to the west. What is the magnitude and direction of the force on the sphere due to the electric field?
(a) Use Gauss's law to prove that the electric field outside any spherically symmetric charge distribution is the same as if all of the charge were concentrated into a point charge. (b) Now use Gauss's law to prove that the electric field inside a spherically symmetric charge distribution is zero if none of the charge is at a distance from the center less than that of the point where we determine the field.
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