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Question-190813




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=((kq)/r^2 )  r^2 =(Q^2 /2)  E_1 =((2kq)/Q^2 )(((−i,j))/( (√2)))  E_2 =((2k(2q))/Q^2 ).(((i,j))/( (√2)))  E_3 =((2k(3q))/Q^2 ).(((i,−j))/( (√2)))  E_4 =((2k(4q))/Q^2 ).(((−i,−j))/( (√2)))  E_(resultante) =(((√2)kq)/Q^2 )[−i+j+2i+2j  +3i−3j−4i−4j)  E_(resul) = ((4(√2)kq)/Q^2 )(−j)
$${E}=\frac{{kq}}{{r}^{\mathrm{2}} } \\ $$$${r}^{\mathrm{2}} =\frac{{Q}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${E}_{\mathrm{1}} =\frac{\mathrm{2}{kq}}{{Q}^{\mathrm{2}} }\frac{\left(−{i},{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{\mathrm{2}} =\frac{\mathrm{2}{k}\left(\mathrm{2}{q}\right)}{{Q}^{\mathrm{2}} }.\frac{\left({i},{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{\mathrm{3}} =\frac{\mathrm{2}{k}\left(\mathrm{3}{q}\right)}{{Q}^{\mathrm{2}} }.\frac{\left({i},−{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{\mathrm{4}} =\frac{\mathrm{2}{k}\left(\mathrm{4}{q}\right)}{{Q}^{\mathrm{2}} }.\frac{\left(−{i},−{j}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$${E}_{{resultante}} =\frac{\sqrt{\mathrm{2}}{kq}}{{Q}^{\mathrm{2}} }\left[−{i}+{j}+\mathrm{2}{i}+\mathrm{2}{j}\right. \\ $$$$\left.+\mathrm{3}{i}−\mathrm{3}{j}−\mathrm{4}{i}−\mathrm{4}{j}\right) \\ $$$${E}_{{resul}} =\:\frac{\mathrm{4}\sqrt{\mathrm{2}}{kq}}{{Q}^{\mathrm{2}} }\left(−{j}\right) \\ $$$$ \\ $$

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