Question #291155

Three point charges q1, q2 and q3 are located at the corners of an equilateral triangle with sides of 10 cm. The electrical forces between them are F12= 5.4 N (attractive), F13= 15 N (repulsive) and F23= 9 N (attractive). Given that q1 is negative, what are the values of q2 and q3?

(Answer: q2= 1.89 μC and q3= -5.28 μC)


1
Expert's answer
2022-01-27T13:23:57-0500

Gives

F12=5.4NF_{12}=5.4N

F13=15NF_{13}=15N

F23=9NF_{23}=9N

Now

F13=kq1q3r2F_{13}=\frac{kq_1q_3}{r^2}

15=9×109×q1×q310215=\frac{9\times10^9\times q_1\times q_3}{10^{-2}}

q1q3=1.66×1011(1)q_1q_3=1.66\times10^{-11}\rightarrow(1)

F12=kq1q2r2F_{12}=\frac{kq_1q_2}{r^2}

5.4=9×109q1q2102q1q2=6×1012(2)-5.4=\frac{9\times10^{9}q_1q_2}{10^{-2}}\\q_1q_2=-6\times10^{-12}\rightarrow(2)

F23=kq2q3r2F_{23}=\frac{kq_2q_3}{r^2}

q2q3=1×1012(3)q_2q_3=-1\times10^{-12}\rightarrow(3)

Equation(2)dividedby(3)Equation (2) divided by(3)

q1q3=6\frac{q_1}{q_3}=-6

q1=6q3q_1=-6q_3

Equation (1) put value


6q32=1.66×1010q3=5.28×106C=5.28μC6q_3^2=-1.66\times10^{-10}\\q_3=-5.28\times10^{-6}C=-5.28\mu C

Equation (3) put q3q_3 Value


5.28×106×q2=1×1012-5.28\times10^{-6}\times q_2=-1\times10^{-12}

q2=1.89×106=1.89μCq_2=1.89\times10^{-6}=1.89\mu C


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