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.

Explanation

`AD = 2cm = 2 times 10^(-2) m`
`CD = 2cm = 2 times 10^(-2) m`
`BD = 2 sqrt 2cm = 2 sqrt 2 times 10^(-2) m`

`E_(A) = (1)/(4 pi varepsilon _(0)) = (Q_(A))/(r^(2)) = 9 times 10^(9) times (4 times 10^(-9))/((2 times 10^(-2))^(2))=9 times 10^(4) NC^(-1)` along AD
`E_(C) = (1)/(4 pi varepsilon _(0)) = 9 times 10^(9) times (4 times 10^(-9))/((2 times 10^(-2))^(2))=9 times 10^(4) NC^(-1)` along CD
`E_(B) = (1)/(4 pi varepsilon _(0)) (Q_(B))/(r^(2)) = 9 times 10^(9) times (4 times 10^(-9))/((2sqrt2 times 10^(-2))^(2))= 4.5times 10^(4) NC^(-1)` along BD
X-Component of the field, `E_(X) = E_(A) + E_(B) cos 45 = 9 times 10^(4) + 4.5 times 10^(4) times 0.707`
`E_(X) = 12.1815 times 10^(4) NC^(-1)`
Y-Component of the field, `E_(Y) =E_(C) + E_(B) cos 45 = 9 times 10^(4) + 4.5 times 10^(4) times 0.707`
`E_(X) = 12.1815 times 10^(4) NC^(-1)`
Magnitude E `=sqrt (E_(X)^(2) + E_(Y)^(2)) = sqrt ((12.1815 times 10^(4))^(2) + (12.1815 times 10^(4))^(2))`
`E= sqrt (2 times(12.1815 times 10^(4))^(2)) = 17.2272 times 10^(4) NC^(-1)`
OR
`E_(AC)= sqrt (E_(A)^(2) + E_(C)^(2)) = sqrt ((9 times 10^(4))^(2) + (9 times 10^(4))^2)`
`E_(AC) = sqrt (2 times (9 times 10^(4))^(2)) = 12.726 times 10^(4) NC^(-1)` along BD
`E = E_(B) + E_(AC) = 4.5 times 10^(4) + 12.726 times 10^(4) = 17.226 times 10^(4) NC^(-1)` along BD