(20%) Problem 2: In the figure, the point charges are located at the corners of
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(20%) Problem 2: In the figure, the point charges are located at the corners of an equilateral triangle 29 cm on a side. ga Ic Otheexpertta.co 25% Part (a) Find the magnitude of the electric field in NC at the location of qa, given that q-8.4 C and q.--4.9 pC . Potential100 cos0 cotan asin0acos0 atan0 acotan sinho Degrees Radians 4 5 6 (2% per attempt) detailed view 1 give up! | 25% Part (b) Find the direction of the electric field at q, in degrees above the negative x-axis with origin at -là 25% Part (c) What is the magnitude of the force in Non qa, given that q.-1.5 nC? 25% Part (d) What is the direction of the force on q, in degrees above the negative x-axis with origin at 0Explanation / Answer
a] Electric field = -(kqb / a^2 cos 60 degree) i+ (kqc / a^2 cos 60 degree) i - (kqb / a^2 sin 60 degree) i - (kqc / a^2 sin 60 degree)
= ( -9e9*8.4e-6/0.29^2 * cos 60 degree + 9e9*-4.9e-6/0.29^2 * cos 60 degree) i + (9e9*8.4e-6/0.29^2 * sin 60 degree + 9e9*-4.9e-6/0.29^2 * sin 60 degree) j
= -711652 i + 324373 j
= sqrt(324373^2+711652^2)
= 782091 N/C
b] direction = arctan(324373/711652) = 24.5 degree
c] Force = qE = 782091*1.5e-9 = 0.001173 N
d] direction = 24.5 degree
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