2 Answers

Rogus · Answer 1 · 2020-01-05T22:27:04+0000

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)

Step-by-step 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 = -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.

Gustavo Berwanger · Answer 2 · 2020-01-09T02:56:19+0000

Answer:

The electric field at x = 0.500 m is 0.02 N/C.

Step-by-step explanation:

Given that,

Point charge at the origin
$q_(1) = 7.00\ nC$

Second point charge
$q_(2)=-250\ nC$ at x = +0.800 m

We calculate the electric field at x = 0.500 m

Using formula of electric field

E=(kq)/(r^2)

The electric field at x = 0.500 m

E=(k*7*10^(-9))/((5)^2)+(k*(-2.50)*10^(-9))/((3)^2)

E=9*10^(9)((7*10^(-9))/(25)-(2.50*10^(-9))/(9))

$E = 0.02\ N/C$

Hence, The electric field at x = 0.500 m is 0.02 N/C.

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