a) Find the magnitude of the electric field, in newtons per coulomb, at the cent
ID: 1878060 • Letter: A
Question
a) Find the magnitude of the electric field, in newtons per coulomb, at the center of the triangular configuration of charges, given that qa = 2.6 nC , qb = -6.2 nC , and qc = 1.3 nC.
b) Find the direction of the electric field in degrees below the right-pointing horizontal (thepositive x-axis).
(8%) Problem 9: In the figure, the point charges are located at the corners of an cquilateral triangle 27 cm on a side da qc gb Otheexpertta.com 50% Part (a) Find the magnitude of the electric field, in newtons per coulomb, at the center of the triangular configuration of charges, given that qa-2.6 nC , qb--6.2 nC , and qc- 13 nC. Grade Summary Deductions Potential 0% 100% cos0 cotanasin)acos0 sin tan() Submissions Attempts remaining: 5 atan)acotan() acotan sinh) 12 3 % per attempt) detailed view cosh)tanh)cotanh) Degrees O Radians Submit Hint I give up! Hints: 1 % deduction per hint. Hints remaining:2 Feedback: deduction per feedback. 1% 50% Part (b) Find the direction of the electric field in degrees below the right-pointing horizontal (the positive x-axis)Explanation / Answer
a] We know that adding equal charges to all the corners will not make a difference, so we add 6.2 nC to all the corners, now the charges are qa = 2.6+6.2 = 8.8 nC, qb = 0, qc = 1.3+6.2 = 7.5 nC
Ex = summation kq/r^2 = 9e9*7.5e-9/[0.27/(sqrt 3)]^2 cos 30 degree = 2405.6 N/C
Ey = summation kq/r^2
= - 9e9*8.8e-9/[0.27/(sqrt 3)]^2 + 9e9*7.5e-9/[0.27/(sqrt 3)]^2 sin 30 degree = -1870 N/C
E = sqrt(2405.6^2+1870^2)
= 3047 N/C
B] direction angle = arctan(1870/2405.6) below x axis
= 37.9 degree below +x axis
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