PLEASE SHOW ALL WORK AND EQUATIONS AND CLEARLY INDICATE ANSWER AND LABEL ANY DIA

Question

PLEASE SHOW ALL WORK AND EQUATIONS AND CLEARLY INDICATE ANSWER AND LABEL ANY DIAGRAMS USED

1) Three identical particles, A, B, and C, each with charge q= 10.00 micro Coulombs, are placed along a circle of radius r = 1.00 m at angles 0, 90, 180 degrees from the positive x axis, respectively. All angles are measured counterclockwise from the positive x axis.

(SHOWN IN DIAGRAM BELOW) (PROBLEM HAS 4 PARTS LABLED A-D)

A) Sketch above qualitatively the net electric feild at the CENTER of the circle and the net Coulomb force on the charge at 90 Degrees.

B) Calculate the net electric field (both magnitude and direction) at the center of the circle. What is the net force (both magnitude and direction) expiernced by the charge at 90 degrees.

C) Calculate the net work done by the feild due to the charge at 0 and 90 degrees to move the charge at 180 degrees to the center of the circle.

D) Now, let all the three charges be placed in the center of a tetrahedron shown below, composed of four triangular faces of equal area. Calculate the net flux through each surface. Each side of the tetrahedon is 0.50 meters.

Explanation / Answer

Part A)

The sketch would be simple. At the center of the circle, draw a vector straight down to represent the net Electric Field. The only field that will be significant is that from the charge at 90obecause the fields from the charges at 0oand 180owill cancel each other out.

For the net force, draw a vector straight up at the charge located at 90o. Similar to the E field at the center, the forces from the charges at 0oand 180owill have x components that cancel, buty components that add.

Part B)

We know that the E field from the 0oand 180ocharges will cancel, so we only need to calculate the field from the 90ocharge.

E = kq/r2

E = (9 X 109)(10 X 10-6)/(12)

E =9 X 104N/C directed straight down

The net force on the 900charge is the addition of the two y components from the other two charges.

The formula for force is F= kqq/d2

We need to find the distance between the charges. The line between the charges on the side to the top is a hypotenuse of a right triangle formed by two side that are the radius of the circle. Therefore, the distance is found by...

d = ?(12+ 12) = 1.414 m, and the angle is obviously 45osince the legs are the same size

Since we need two y components, we have

F = 2(kqq/d2)(cos 45)

F = (2)(9 X 109)(10 X 10-6)2(cos 45)/(1.414)2

F =.637 N directed upward

Part C

The work done is found by the change in potential such as W = q?V

We can find the total potential at the location of the charge at 180oand the potential at the center caused by the other two charges. Potential is a scalar quantity so it is added algebraically,

V = kq/r

At the 180olocation, one charge is 1.414 m away, and the other is 2 m away

V = (9 X 109)(10 X 10-6)/1.414 + (9 X 109)(10 X 10-6)/2 = 1.09 X 105V

At the center, the distance is 1 m for both charges, so...

V = 2(kq/r) = (2)(9 X 109)(10 X 10-6)/1 = 1.80 X 105V

Finally W = q?V = (10 X 10-6)(1.80 X 105- 1.09 X 105)

W = .71 J

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