Physics Case: Electric Charges and Forces.

[Firu]

Homework Statement

Four punctual charges are located as shown in the figure below. A, B, and C are located at the vertices of an equilateral triangle. The values for these charges are A = +1.4microculombs, B = 4.5microculombs, C = -6.7microculombs and D = +7.1microculboms. THe length of the triagnle's base is shown in the figure. Calculate the net force experienced by charge D (magnitude and direction from the X - Axis). D is on the center of the triangle.

--------------A-----------------------------D---------------------C-------------B---------

Distance between C -B = 7.1nm

Homework Equations

F= K*q1*q2/R^2

The Attempt at a Solution

Converted charges to culombs, so i got
A = 1.4x10^-5
B = 4.5x10^-5
C = -6.7x10^-5

Then converted the distance to meters so got:
7.8x10^-8

Used equilateral triangle rules to get the distance between A - D, B - D, and C -D (which is the same) and got R = 4.5x10-8.

K = 9x10^9.

Used the F formula on each pair (A -D, B -D, C - D) using

F = k*q1*q1/R^2.

On every formula q1 = D (7.1x10^-5) and q2 was A, B or C.
R^2 is: 2.025x10-15.

Forces I got were.

A - D = 4.41x10^15 = F1
B - D = 1.42x10^16 = F2
C - D = -2.1x10^16 = F3

Net force =
F1+F2+F3

Net Force = -2.39x10^15.
Moving towards C according to the diagram.

Is this correct?

Thanks.

 

[mark726]

Yes, your solution is correct. The net force experienced by charge D is -2.39x10^15 N, directed towards C.