Consider the following figure. (The four charges are at the corners of a square.

Question

Consider the following figure. (The four charges are at the corners of a square. Take q = 1.03

Consider the following figure. (The four charges are at the corners of a square. Take q = 1.03 muC and Q = 2.15 muC.) In the figure, what are the magnitude and direction of the resultant force on q1? (Take q1 as the origin of the coordinate system and measure the angle counterclockwise from the positive x-axis, which is directed towards the right.) (b) What is the resultant force on the center of mass of the four charges?

Explanation / Answer

Part A)

For the force from the right

F = kqq/r^2

F = (9 X 10^9)(1.03 X 10^-6)(2.15 X 10^-6)/(.1)^2

F = 1.99 N to the left

From the force below

F = (9 X 10^9)(1.03 X 10^-6)(1.03 X 10^-6)/(.1)^2

F = .955 N straight down

From the force at the diagonal

F = (9 X 10^9)(1.03 X 10^-6)(2.15 X 10^-6)/(.1414)^2

F = .997 N at a 45 degree angle

The components are (.997)(cos 45) = .705 N down and to the right

Net x = 1.99 - .705 = 1.285 N to the left

Net y = .955 + .705 = 1.66 down

Net force = sqrt[(1.285)^2 + (1.66)^2]

Net Force = 2.10 N

The angle...

tan(angle) = 1.66/1.285

angle = 52.3 degrees

Add 180 for the direction from the positive x...

180 + 52.3 = 232.5

So, in summary

2.10 N at 232.5 degrees

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