Answer to Question #123559 in Electricity and Magnetism for Jaykay

Question #123559
a square with sides 10cm has three charges on it edges, two are+5mC and one is -3mC. draw a well labelled diagram to illustrate the question and compute the total field at the vacant edge due to the other charges
1
Expert's answer
2020-06-23T12:57:25-0400

E1=kq1r12=910931030.12=2.7109V/mE_1=k\frac{q_1}{r_1^2}=9\cdot10^9\cdot \frac{3\cdot10^{-3}}{0.1^2}=2.7\cdot10^9V/m


E2=kq2r22=910951030.12=4.5109V/mE_2=k\frac{q_2}{r_2^2}=9\cdot10^9\cdot \frac{5\cdot10^{-3}}{0.1^2}=4.5\cdot10^9V/m


E3=kq3r32=91095103(0.12)2=2.25109V/mE_3=k\frac{q_3}{r_3^2}=9\cdot10^9\cdot \frac{5\cdot10^{-3}}{(0.1\sqrt{2})^2}=2.25\cdot10^9V/m


Ex=E3cos45°+E2=(2.25cos45°+4.5)109=6.1109V/mE_x=E_3\cos45°+E_2=(2.25\cos45°+4.5)\cdot10^9=6.1\cdot10^9V/m


Ey=E3sin45°+E1=(2.25sin45°+2.7)109=1.1109V/mE_y=-E_3\sin45°+E_1=(-2.25\sin45°+2.7)\cdot10^9=1.1\cdot10^9V/m


E=Ex2+Ey2=1096.12+1.12=6.2109V/mE=\sqrt{E_x^2+E_y^2}=10^9\cdot\sqrt{6.1^2+1.1^2}=6.2\cdot10^9V/m











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