Respuesta :
Answer:
The electric field at x = 0.500 m is 0.02 N/C.
Explanation:
Given that,
Point charge at the origin[tex]q_{1} = 7.00\ nC[/tex]
Second point charge[tex]q_{2}=-250\ nC[/tex] Â at x = +0.800 m
We calculate the electric field at x = 0.500 m
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
The electric field at x = 0.500 m
[tex]E=\dfrac{k\times7\times10^{-9}}{(5)^2}+\dfrac{k\times(-2.50)\times10^{-9}}{(3)^2}[/tex]
[tex]E=9\times10^{9}(\dfrac{7\times10^{-9}}{25}-\dfrac{2.50\times10^{-9}}{9})[/tex]
[tex]E = 0.02\ N/C[/tex]
Hence, The electric field at x = 0.500 m is 0.02 N/C.
Answer: The electric field at x = 0.5 m is equal to 1.96 N/C, and the direction is in the postive x-axis (to the rigth)
Explanation:
I will use the notations (x, y, z)
the first particle is located at the point (0m, 0m, 0m) and has a charge q1 = 7.00 nC
the second particle is located at the point (0.8m, 0m, 0m) and has a charge q2 = Â -2.50 nC
Now, we want to find the electric field at the point (0.5m, 0m, 0m)
First, we can see that we only work on the x-axis, so we can think about this problem as one-dimensional.
First, the electric field done by a charge located in the point x0 is equal to:
E(x) = Kc*q/(x - x0)^2
where Kc is a constant, and it is Kc = 8.9*10^9 N*m^2/C^2
then, the total magnetic field will be equal to the addition of the magnetic fields generated by the two charges:
E(0.5m) = Kc*q1/0.5m^2 + Kc*q2/(0.5m - 0.8m)
E(0.5m) = Kc*(7.0nC/(0.5m)^2 - 2.5nC/(0.3m)^2)
E(0.5m) = Kc*(0.22nC/m^2)
now, remember that Kc is in coulombs, so we must change the units from nC to C
where 1nC = 1*10^-9 C
E(0.5m) = (8.9*10^9 N*m^2/C^2)*(0.22x10^-9C/m^2) = 1.96 N/C
the fact that is positive means that it points in the positve side of the x-axis.
