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Three charges each equal to +4nC are placed at the three comers of a square of side 2 cm. Find the electric field at the fourth corner. |
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Answer» Solution :Diagonal BD=(`sqrt2`) (side) i.e. `BD=2sqrt2xx10^(-2)`m we know that ,`E=(1/(4pi in_0))(q/r^2)` HENCE, `E_1=(9xx10^9xx4xx10^(-9))/((2xx10^(-2))^2)=(36xx10^4)/4` `=9xx10^4 Vm^(-1)` ALONG AD `E_2=(9xx10^9xx4xx10^(-9))/((2sqrt2xx10^(-2))^2)=(36xx10^(4))/8` `E_3=(9xx10^9xx4xx10^(-9))/((2xx10^(-2))^2)=9xx10^4Vm^(-1)`along CD RESULTANTOF `E_1` and `E_3` , `E.=sqrt(E_1^2 +E_3^2 +2E_1E_3 cos 90^@)` & `E_1 =E_3`. `therefore E.=sqrt2E_1 = 1.414xx9xx10^4` `=12.726xx10^4 Vm^(-1)`, along `E_2` Hene resultantfieldat D =`E^1 +E_2 =(12.726+4.5)xx10^4 Vm^(-1)` `=17.226xx10^4` `=1.723xx10^5 Vm^(-1)` at `45^@`along `vecE_2` 
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