(a) Three point charges, A = 1.70 µC, B = 6.90 µC, and C = -3.80 µC, are located

Question

(a) Three point charges, A = 1.70 µC, B = 6.90 µC, and C = -3.80 µC, are located at the corners of an equilateral triangle as in the figure above. Find the magnitude and direction of the electric field at the position of the 1.70 µC charge.

magnitude    
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N/C direction    
Your response differs from the correct answer by more than 10%. Double check your calculations.° (counterclockwise from the +x-axis)

Explanation / Answer

here

by using the formula of electric field = k * q / r^2

the electric field at A from B is at 240degrees with magnitude = k * 6.9 * 10^-6 / 0.5^2 = 248056 N/C

then

the electric field at S from C is at 0 degrees with magnitude = k * 3.8 * 10^-6 / 0.5^2 = 136800 N/C

the sum horizontal components = 248056 * cos(240) + 136800* cos0 = 12772

then

sum of vertical components = 248056 * sin240 + 136800 * sin0 = -214823

so the magnitude is = sqrt( 12772^2 + 214823^2) = 215202 N/C

afterthis the direction is

theta = tan^-1(-215202/12772) = -86.6deg

theta = 360 - 86.6 = 273.4 deg

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