Coulomb's law and electric field intensity
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The electric field from any single line charge always points directly away from the line charge, is proportional to the linear charge density, and falls off like $\frac{1}{r}$
In your problem, at any point of interest, you will have an electric field vector that has a component on the yz plane from the line charge on the x axis, and a component on the xz plane from the line charge on the y axis.
The total magnitude will be $E=\sqrt{E_{yz}^2 + E_{xz}^2}$
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DONGHUN
Updated on October 21, 2022Comments

DONGHUN 7 months

DONGHUN about 7 yearsI understand whole your comment, but I don't catch why R_x= 4a_z still...