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Can these two vector fields be possible electric fields? $$ egin{aligned} &vec{E}=left(frac{1}{x+2 y+3 z} ight) hat{i}+left(frac{2}{x+2 y+3 z} ight) hat{j}+lef...

Question

Can these two vector fields be possible electric fields? $$ egin{aligned} &vec{E}=left(frac{1}{x+2 y+3 z} ight) hat{i}+left(frac{2}{x+2 y+3 z} ight) hat{j}+left(frac{3}{x+2 y+3 z} ight) hat{k} \ &vec{E}=left(frac{2}{x+2 y+3 z} ight) hat{i}+left(frac{1}{x+2 y+3 z} ight) hat{j}+left(frac{4}{x+2 y+3 z} ight) hat{k} end{aligned} $$ If the vector fields are indeed possible electric fields, show explicitly that the work done in moving a unit charge from origin to the point $(1,1,1)$ via the f

Can these two vector fields be possible electric fields? $$ egin{aligned} &vec{E}=left(frac{1}{x+2 y+3 z} ight) hat{i}+left(frac{2}{x+2 y+3 z} ight) hat{j}+left(frac{3}{x+2 y+3 z} ight) hat{k} \ &vec{E}=left(frac{2}{x+2 y+3 z} ight) hat{i}+left(frac{1}{x+2 y+3 z} ight) hat{j}+left(frac{4}{x+2 y+3 z} ight) hat{k} end{aligned} $$ If the vector fields are indeed possible electric fields, show explicitly that the work done in moving a unit charge from origin to the point $(1,1,1)$ via the following two paths are same: (a) along a straight line and (a) along the $x$ -axis to the point $(1,0,0)$ and then parallel to the $y$ -axis to $(1,1,0)$ and then parallel to $z$ -axis to $(1,1,1)$.



Answers

Two point charges $Q$ and $+$q (where q is positive) produce the net electric field shown at point $P$ in $\textbf{Fig. E21.36.}$ The field points parallel to the line connecting the two charges. (a) What can you conclude about the sign and magnitude of $Q$? Explain your reasoning. (b) If the lower charge were negative instead, would it be possible for the field to have the direction shown in the figure? Explain your reasoning.


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