Answer:
E =+823.12N/C.
Explanation:
The electric field [tex]E[/tex] at point [tex]P[/tex] is the sum of electric field [tex]E_1[/tex] due to [tex]q_1[/tex] and [tex]E_2[/tex] due to [tex]q_2[/tex] at [tex]P[/tex]:
[tex]E = E_1+E_2[/tex].
The distance from [tex]P[/tex] to [tex]q_1[/tex] is [tex]R_1 = 0.150m+0.250m = 0.400m[/tex]; therefore, the electric due to it is
[tex]E_1 = k\dfrac{q_1}{R_1^2}[/tex]
[tex]E_1 = 9*10^9*\dfrac{+6.39*10^{-9}C}{(0.400m)^2}[/tex]
[tex]E_1 = +359.44N/C.[/tex]
And the distance from [tex]P[/tex] to [tex]q_2[/tex] is [tex]R_2 = 0.250m[/tex]; therefore,
[tex]E_2 = k\dfrac{q_2}{R_2^2}[/tex]
[tex]E_2 = 9*10^9*\dfrac{+3.22*10^{-9}C}{(0.250m)^2}[/tex]
[tex]E_2 = +463.68N/C[/tex]
Hence, the total electric field at point [tex]P[/tex] is
[tex]E = E_1+E_2[/tex]
[tex]E = +359.44N/C+463.68N/C[/tex]
[tex]\boxed{E =+823.12N/C.}[/tex]
where the plus sign indicates that the the electric field points to the right.
