Respuesta :
Answer:
[tex]E_x = -5.93 * 10^6[/tex]
Step-by-step explanation:
Parameters given:
Point charge at the center of the sphere, [tex]q_1[/tex] : [tex]-7.7 * 10^{-6}[/tex]μC
Charge of the sphere, [tex]q_2[/tex] : [tex]2.6 * 10^{-6}[/tex]μC
Distance between [tex]q_1[/tex] and the point of consideration = [tex]8.8 * 10^{-2} m[/tex]
Distance between [tex]q_2[/tex] and the point of consideration = [tex]8.8 * 10^{-2} m[/tex]
Electric field is given as
[tex]E_x = \frac{kq}{r^2}[/tex]
where
k = Coulombs constant;
q = electric charge;
r = distance between charge and point of consideration.
The net electric field at that point is the sum of the electric field due to [tex]q_1[/tex] and [tex]q_2[/tex], i.e.:
[tex]E_x = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}[/tex]
Since k is the same and the distance, r is also the same, then:
[tex]E_x =\frac{k}{r^2} ( q_1 + q_2)[/tex]
=> [tex]E_x =\frac{9 * 10^9}{(8.8 * 10^{-2})^2} [ (-7.7 * 10^{-6}) + (2.6 * 10^{-6})][/tex]
=> [tex]E_x = 1.162 * 10^{12} * -5.1 * 10^{-6}\\\\\\E_x = -5.93 * 10^6[/tex]
The electric field along the x axis, [tex]E_x = -5.93 * 10^6[/tex]
