Question Number 190813 by Spillover last updated on 12/Apr/23
Answered by manxsol last updated on 12/Apr/23
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Commented by manxsol last updated on 13/Apr/23
$${E}=\frac{{kq}}{{r}^{\mathrm{2}} } \\ $$$${r}^{\mathrm{2}} =\frac{{Q}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${E}_{\mathrm{1}} =\frac{\mathrm{2}{kq}}{{Q}^{\mathrm{2}} }\frac{\left(−{i},{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{\mathrm{2}} =\frac{\mathrm{2}{k}\left(\mathrm{2}{q}\right)}{{Q}^{\mathrm{2}} }.\frac{\left({i},{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{\mathrm{3}} =\frac{\mathrm{2}{k}\left(\mathrm{3}{q}\right)}{{Q}^{\mathrm{2}} }.\frac{\left({i},−{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{\mathrm{4}} =\frac{\mathrm{2}{k}\left(\mathrm{4}{q}\right)}{{Q}^{\mathrm{2}} }.\frac{\left(−{i},−{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{{resultante}} =\frac{\sqrt{\mathrm{2}}{kq}}{{Q}^{\mathrm{2}} }\left[−{i}+{j}+\mathrm{2}{i}+\mathrm{2}{j}\right. \\ $$$$\left.+\mathrm{3}{i}−\mathrm{3}{j}−\mathrm{4}{i}−\mathrm{4}{j}\right) \\ $$$${E}_{{resul}} =\:\frac{\mathrm{4}\sqrt{\mathrm{2}}{kq}}{{Q}^{\mathrm{2}} }\left(−{j}\right) \\ $$$$ \\ $$