The diagram below shows four charges at the corners of a square of sides d = 1.4

Question

The diagram below shows four charges at the corners of a square of sides d = 1.40 m. Here, q1 = q2 =-q and q3 = 2.0q, where q = 3.00 nC and 93 is located at the origin. What is the force on the fourth charge q4-3.50 nC? Assume that the +x axis is directed to the right and the +y axis is directed up. Express your answer in vector form. Do not enter units in your expression F=13.6e-8 Draw a diagram and label all the forces appropriately. What is the magnitude of the force exerted by each of the three charges on q4? What are the directions of these forces? N 92 94 73 41

Explanation / Answer

q=3nC=3*10-9 C

q1=q2=-q=-3*10-9 C

q3=2q=2*3*10-9 =6*10-9 C

q4=3.5*10-9 C

Force on q4 due to q1 = 9*109*3*10-9*3.5*10-9/1.42 = 48.2*10-9 N (negative y direction)=-48.2*10-9 j N

Force on q4 due to q2 = 9*109*3*10-9*3.5*10-9/1.42 = 48.2*10-9 N (negative x direction)=-48.2*10-9 i N

Force on q4 due to q3 = 9*109*6*10-9*3.5*10-9/[1.4*srtq(2)]2 = 48.2*10-9 N along the digonal at 45o anticlockwise from +xaxis = 34.09*10-9 i + 34.09*10-9 j N

resultant force on q4 = -48.2*10-9 j-48.2*10-9 i+34.09*10-9 i + 34.09*10-9 j =-14.11*10-9 i - 14.11*10-9 j N

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