Answer to Question #300744 in Electricity and Magnetism for Ron

Question #300744

A point charge π‘ž1 = +6.00 π‘₯ 10βˆ’9𝐢 is at the point π‘₯ = 0.800 π‘š, 𝑦 = 0.600 π‘š and a second point charge π‘ž2 = βˆ’2.00 π‘₯ 10βˆ’9𝐢 is at the point π‘₯ = 0.800 π‘š, 𝑦 = 0 π‘š. Calculate the magnitude and direction of the resultant electric field at the origin due to these charges.

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Expert's answer
2022-02-21T14:56:46-0500

E1=kq1r12=54.0 NC,E_1=\frac{kq_1}{r_1^2}=54.0~\frac NC,

E2=kq2r22=28.1 NC,E_2=\frac{kq_2}{r_2^2}=28.1~\frac NC,

E=E12+E22+2E1E2sin⁑φ=35.7 NC.E=\sqrt{E_1^2+E_2^2+2E_1E_2\sin\varphi}=35.7~\frac NC.


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