Question

Three electric charges are located as shown in the diagram, with--7.3.pC,6.8.C, 6id, b 0.61.m, and c 0.65.m. What is the electric force on charge q due to charges q, and q a0.35.m, c .SSm -63 1 a 42 11. magnitude 403153113:I .1x9 a. 1.025,N b. 0.8985 N c. 1.172.N d. 0.9356.N e. 1.796,N . 1.618.N O 13.5111 a. 12. direction (as rotation angle) a 10.64 b. 7.229 c. 14.79° d. 7.287 e 21.21 . 13.07

Answer 1

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We know that electrostatic force is given by:

F = k*Q1*Q2/R^2

Now Force on q due to q1 will be repulsive (In north of east direction) as both charges have same sign, Now this force will be

F1 = k*q*q1/r1^2

q = charge on q = -7.3 uC = -7.3*10^-6 C

q1 = charge on q1 = -6.8 uC = -6.8*10^-6 C

r1 = distance between q and q1 = sqrt (0.35^2 + 0.65^2) = 0.738 m

So,

F1 = 9*10^9*7.3*10^-6*6.8*10^-6/0.738^2 = 0.8202 N

Now x-component of this force will be

F1x = F1*cos theta

cos theta = 0.35/0.738, So

F1x = 0.8202*0.35/0.738 = 0.389 N

F1y = F1*sin theta

sin theta = 0.65/0.738, So

F1y = 0.8202*0.65/0.738 = 0.722 N

Similarly Force on q due to q2 will be attractive and towards South of east direction, So

F2 = k*q*q2/r2^2

q2 = 8.6 uC = 8.6*10^-6 C

r2 = sqrt (0.61^2 + 0.65^2) = 0.891 m

So,

F2 = 9*10^9*7.3*10^-6*8.6*10^-6/0.891^2 = 0.712 N

Now

F2x = F2*cos theta = 0.712*0.61/0.891 = 0.487 N

F2y will be negative because it's in south direction. So

F2y = -F2*sin theta = -0.712*0.65/0.891 = -0.519 N

Net Force will be

F = F1 + F2

F = (F1x + F2x) i + (F1y + F2y) j

F = (0.389 + 0.487) i + (0.722 - 0.519) j

F = 0.876 i + 0.203 j

Magnitude of net force will be

|F| = sqrt (0.876^2 + 0.203^2) = 0.899 N

Correct option is B.

2.

Direction is given by:

Direction = arctan (Fy/Fx)

Direction = arctan (0.203/0.876) = 13.05 deg

Correct option is F.

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