Answer:
[tex]E_n=34,467,075.42\ N/C[/tex]
Step-by-step explanation:
Electric Field
The electric field produced by a point charge Q at a distance d is given by
[tex]\displaystyle E=K\cdot \frac{Q}{d^2}[/tex]
Where
[tex]K = 9\cdot 10^9\ Nw.m^2/c^2[/tex]
The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.
Let's calculate the electric fields of each charge at the given point. The first charge [tex]q_1=+6.24\mu C=6.24\cdot 10^{-6}C[/tex] is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:
[tex]\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}[/tex]
[tex]E_1=37,888,345.42\ N/C[/tex]
Similarly, for [tex]q_2=-9.55\mu C=-9.55\cdot 10^{-6}C[/tex], the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right:
[tex]\displaystyle E_2=9\cdot 10^9\cdot \frac{9.55\cdot 10^{-6}}{0.1585^2}[/tex]
[tex]E_2=3,421,270\ N/C[/tex]
Since E1 and E2 are opposite, the net field is the subtraction of both
[tex]E_n=37,888,345.42\ N/C-3,421,270\ N/C[/tex]
[tex]\boxed{E_n=34,467,075.42\ N/C}[/tex]
