2. Find the magnitude and direction of the net electric field at the center of the triangle in the figure. The triangle is equilateral, so all sides are the same 3.0cm and all three interior angles are 60°. A second, right triangle with base equal to ½ the base of the large triangle has been added to assist -1nC you. The hypotenuse of the small right triangle bisects the interior angle of the larger triangle. -1nc +2nC 3.0cm ক শ প শদ আা

2. Find the magnitude and direction of the net electric field at the center of the triangle in the figure. The triangle is equilateral, so all sides are the same 3.0cm and all three interior angles are 60°. A second, right triangle with base equal to ½ the base of the large triangle has been added to assist -1nC you. The hypotenuse of the small right triangle bisects the interior angle of the larger triangle. -1nc +2nC 3.0cm ক শ প শদ আা

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter18: Electric Charge And Electric Field

Section: Chapter Questions

Problem 46PE: (a) Using the symmetry of the arrangement, determine the direction of the electric field at the...

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Two equal and opposite charges of magnitude Q are located on the x-axis at the points +a and a , as shown below. What is the net flux due to these charges through a square surface of side 2a that lies in the yz-plane and is centered at the origin? (Hint: Deter-nine the flux due to each charge separately, then use the principle of superposition. You may be able to make a symmetry argument.)
ne electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude 3 N/C and directed outward from the axis of the rod. (a) How much charge per unit length exists on the rod? (b) What would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly?
In which of the following contexts ran Gausss law not be readily applied to find the electric field? (a) near a long, uniformly charged wire (b) above a large, uniformly charged plane (c) inside a uniformly charged ball (d) outside a uniformly charged sphere (e) Gausss law can be readily applied to find the electric field in all these contexts.
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(a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.53, given that qa= 1.00C and qc=qd= +1.00 C. (b) Calculate the magnitude of the electric field at the location of q, given that the square is 5.00 cm on a side.
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Example 23.3 derives the exact expression for the electric field at a point on the axis of a uniformly charged disk. Consider a disk of radius R = 3.00 cm having a uniformly distributed charge of +5.20 C. (a) Using the result of Example 23.3, compute the electric field at a point on the axis and 3.00 mm from the center. (b) What If? Explain how the answer to part (a) compares with the field computed from the near-field approximation E = /20. (We derived this expression in Example 23.3.) (c) Using the result of Example 23.3, compute the electric field at a point on the axis and 30.0 cm from the center of the disk. (d) What If? Explain how the answer to part (c) compares with the electric field obtained by treating the disk as a +5.20-C charged particle at a distance of 30.0 cm.
The charge unit length on the thin shown below is .What is the force on the point charge q? Solve this problem by first considering the electric force dF on q due to a small segment dx of the rod, which contains charge dx . Then, find the net force by integrating dF over the length of the rod.
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