Answer:[tex]2.88\times 10^3[/tex]
Explanation:
Given
Charge on A is [tex]4 \mu C[/tex] at (-3,0)
Charge on B is [tex]-4 \mu C[/tex] at (3,0)
Electric Field at a distance d is given by [tex]E=\frac{kQ}{d^2}[/tex]
Thus Electric filed at (0,4) due to A is
[tex]E_A=\frac{kQ_A}{d^2}=\frac{1}{4\pi \epsilon }\frac{Q_A}{d^2}[/tex]
Here [tex]d =\sqrt{3^2+4^2}=5[/tex]
[tex]E_A=\frac{9\times 10^9\times 4\times 10^{-6}}{5^2}[/tex]
[tex]E_A=1.44\times 10^3 N\C (Away\ From\ Q_A)[/tex]
Similarly [tex]E_B[/tex] is [tex]1.44\times 10^3 N\C (towards Q_B)[/tex]
Electric Field is at an angle [tex]\theta [/tex]which is given by
[tex]tan\theta =\frac{4}{3}[/tex]
Thus [tex]E_Asin\theta [/tex]and [tex]E_Bcos\theta [/tex]component will cancel out each other and Cos component will add up
[tex]E_{net}=E_Acos\theta +E_Bcos\theta [/tex]
[tex]E_{net}=2Ecos\theta [/tex]
[tex]E_{net}=2\times 1.44\times 10^3 N\C[/tex] (towards positive x axis )
