Answer:
[tex]\vec{E} = 1.334\hat{i} + 2.668\hat{j} + 2.668\hat{k}[/tex] N/C
Solution:
Charge, Q = [tex]4\mu C = 4\times 10^{- 6} C[/tex]
Coordinates (x, y, z) are (1, 2, 2)
Now, we know that the Electric field at any point in the Cartesian plane is given by:
[tex]\vec{E} = \frac{1}{4\pi epsilon_{o}}\times \frac{Q}{|d|^{2}}\hat{d}[/tex] (1)
where
[tex]\frac{1}{4\pi epsilon_{o}} = 9\times 10^{9} N-m^{2}/C^{2}[/tex]
d = distance of the charge from the point
[tex]\hat{d} = \hat{i} + 2\hat{j} + 2\hat{k}[/tex]
[tex]|d| = \sqrt{1^{2} + 2^{2} + 2^{2}[/tex]
Now, from eqn (1):
[tex]\vec{E} = 9\times 10^{9}\times \frac{4\times 10^{- 6}}{(\sqrt{1^{2} + 2^{2} + 2^{2})^{2}}}\times (\hat{i} + 2\hat{j} + 2\hat{k})[/tex]
[tex]\vec{E} = 1.334(\hat{i} + 2\hat{j} + 2\hat{k})[/tex] N/C
[tex]\vec{E} = 1.334\hat{i} + 2.668\hat{j} + 2.668\hat{k}[/tex] N/C
