I am having trouble with this problem. I tried using pythagoreon theorum to get

Question

I am having trouble with this problem. I tried using pythagoreon theorum to get the missing sides and using inverse sin and inverse cos to get missing angles. Then using the columbs law for each listed charge and finding the sum of all charges. My professor said the correct answer should be 7.81 x 103 N/C at 10.6 degrees cw from the -x axis. I cannot get my numbers to work. Can you show me a detailed version on what to do for this problem?

Two charged objects (Q1=-9.0 nC and Q2=+15.0 nC) are fixed in place on the x-axis. One charge is 12.0 cm to the right of the origin and the other is 12.0 cm to the left of the origin. Find the net electric field (magnitude and direction) due to Q1 and Q2 at point P, located on the y-axis 9.0 cm from the origin.

Explanation / Answer

The distance from each of the charges to point P is 15 cm or 0.15 m from the Pythagorean theorem. To calculate the field you need the individual components from each charge.

From Q1, E = kq/r2 = (8.988 * 109 Nm2/C2)(-9 * 10-9 C)/(0.15 m)2 = -3595.2 N/C. Since the charge is negative, the field vector points toward the charge so both components are negative.

E1x = (3595.2)(12/15) = -2876.16 N/C

E1y = (3595.2)(9/15) = -2157.12 N/C

For Q2, E = (8.988 * 109 Nm2/C2)(15 * 10-9 C)(0.15 m)2 = 5992 N/C. This time the x component is negative and the y component is positive.

E2x = -(5992)(12/15) = -4793.6 N/C

E2y = (5992)(9/15) = 3595.2 N/C

Adding the x and y components together, the final field is Ex = -7669.76, Ey = 1438.08. The magnitude of the total field is the square root of the sum of the squares of the two components, or 7803 N/C = 7.8 * 103 N/C and the angle is the arctangent of Ey/Ex or 10.6 degrees from the x-axis in quadrant II.

I agree with your professor except for a minor rounding error, he said 7.81 * 103 N/C and I show 7.803 * 103.

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