Ap Coulomb S Law

Purple Squirrel and Blue Angel

READ FIRST: THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS. Future explanations will be in the same way as number one was done. Due to time limiting factors, questions two and three have been typed in a much quicker “format”. 1. Three charges are fixed to an x y coordinate system. A charge of +18 µC is on the y axis at y = +3.0 m. A charge of -8 µC is at the origin. Lastly, a charge of +48 µC is on the x axis at x = +3.0 m. Determine the magnitude and direction of the net electrostatic force on the charge at x = +3.0 m. Specify the direction relative to the -x axis. Solution: Part A: Force 1 = +18uC

Force 2 = -8 uC

Force 3 = +48uC

The net elastic force on charge 3 at x = +3.0m is the sum of all forces affecting it. There are two charges acting on charge 3 (+18uC and -8uC). Coulomb’s Law can be used to find the magnitude of the two charges. To find the force on F13: The radius is found by the Pythagorean Theorem R2 = x2 + y2 R2= (3)2 + (3)2 R2= 18 R=-\/18| F13 = (kq1q3)/r2 (Note: The charges are given in micro Coulombs, multiply by 10-6 to change to Coulombs) F13 = ((9x109) x (18 x 10-6) x (48 x 10-6))/ 18 = .432 N

Since F13 is a vector, it has x and y components which can be found by multiplying by the angle of 45 degrees. (Charges 1 and 3 are equal distance from the origin so the angle is 45)

(*Note: The charges in F13 are both positive so they repel each other. Thus the direction of charge 3 is down and right so the x component is positive and the y component is negative) F13x = .432 x cos45 = .305N F13y = .432 x cos45 = -.305N To find the force on F23: The distance between F23 is 3.0m so the radius is 3.0m F23 = kq1q3/r2 F23 = (9x109) x (8 x 10-6) x (48x10-6) / (3)2 = .384 N Since F23 lies on the x-axis F23 will only have a x component

(*Note: F23 are negative and positive so they will attract one another. Thus the direction of charge 3 is to the left so the x component is negative) F23x = -.384N Adding up the x and y components we get: F3x = F13x + F23x = (.305N) + (-.384N) = -.079N

F3y = F13y = -.305N We can now find the magnitude of force 3 by the Pythagorean Theorem (F3)2 = (x)2 + (y)2 (F3) = (-.079) 2 + (-.305N) 2 2

F3 = .315 N Part B: The angle can be found by tangent inverse of sine (y) divided by cosine (x). tan-1 (.305/.079) = 75.478° Part C: The direction is below the x-axis since the y component is negative.

2. An equilateral triangle has sides of 0.13 m. Charges of -8.6, +8.0, and +1.7 µC are located at the corners of the triangle. Find the magnitude of the net electrostatic force exerted on the 1.7-µC charge.

(Do not attempt to convert the charges from micro Coulombs to Coulomb as it is already done)

q1 = -8.6µC q2 = +8.0µC q3 = +1.7 µC (*Note the charges must be listed in the order they are given)

Step 1 *Multiply .009 times charge 1 *Multiply the answer you just got times charge 3 *Divide the answer you just got by .0169 Step 2 *Multiply step 1 times the cosine of 60 Step 3 *Multiply step 1 times the sine of 60 Step 4 *Multiply .009 times charge 2 *Multiply the answer you just got times charge 3 *Divide the answer you just got by .0169 Step 5 *Subtract step 2 away from step 4 Step 6 *Square step 5 Step 7 *Square step 3 *Add step 6 and step 7 together *Take the square root of the answer you just got 3.*Multiply the point charge by 2 *Divide the number you just got by the cos of 45 *Drop the negative sign !! THESE EXPLANATIONS AND SHORT STEPS BROUGHT TO YOU BY:

Purple Squirrel & Blue Angel THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION.

THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS. Future explanations will be in the same way as number one was done. Due to time limiting factors, questions two and three have been typed in a much quicker “format”.