Four charges are placed at the corners of a square. The side of the square is a
ID: 2282923 • Letter: F
Question
Four charges are placed at the corners of a square. The side of the square is a = 0.33 m.
a) If the charge are identical and positive q = 53.0 x 10-9 C what would be the magnitude of the force acting on each of the charges? Answer in Newtons.
b) We change the charges by q1 = 53.0 x 10-9 C, q2 = 49.0 x 10-9 C, q3 = -55.0 x 10-9 C, q4 = -93.0 x 10-9 C. What would be the magnitude of the force acting on charge q3? Answer in Newtons.
c) We place the same charges at the corners of a rectangle of sides a and b (instead of a square). If b=2a/3 what would be the magnitude of the force acting on charge q3? Answer in units of Newtons.
Explanation / Answer
a) on Q1
due to Q2, Fq2=K*Q1*Q2/a2
= 9*109*53*10-9*53*10-9/.332
= 2.32*10-4<180 N
due to Q3, Fq2=K*Q1*Q3/a2
= 9*109*53*10-9*53*10-9/.332
= 2.32*10-4<90 N
Fq3=K*Q1*Q3/2*a2
= 9*109*53*10-9*53*10-9/2*.332
= 1.16*10-4<135 N
Total Force acting on Q1 = Fq3+Fq2+Fq4= 4.44*10-4<135 N
magnitude = 4.44*10-4 N same force will act on each charge
b) on Q3
due to Q2, Fq2=K*Q3*Q2/a2
= 9*109*49*10-9*55*10-9/2*.332
= 1.11*10-4<45 N
due to Q1, Fq1=K*Q1*Q3/a2
= 9*109*55*10-9*53*10-9/.332
= 2.41*10-4<90 N
due to Q4, Fq4=K*Q4*Q3/a2
= 9*109*55*10-9*93*10-9/.332
= 4.23*10-4<180 N
Total Force acting on Q3 = Fq1+Fq2+Fq4= 4.698*10-4<137 N
magnitude = 4.698*10-4 N
c) on Q3 b=2a/3 = .22
due to Q2, Fq2=K*Q3*Q2/(a2+b2)
= 9*109*49*10-9*55*10-9/(.332+.222)
= 1.54*10-4<33.69 N
due to Q1, Fq1=K*Q1*Q3/b2
= 9*109*55*10-9*53*10-9/.222
= 5.42*10-4<90 N
due to Q4, Fq4=K*Q4*Q3/a2
= 9*109*55*10-9*93*10-9/.332
= 4.23*10-4<180 N
Total Force acting on Q3 = Fq1+Fq2+Fq4= 6.93*10-4<115 N
magnitude = 6.03*10-4 N
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