A particle (charge = +4.0 nC) is located on the x-axis at the point x = -20.0 m, and a second particle (charge = -10.0 nC) is placed on the x-axis at x=+8.0 m. What is the magnitude of the total electrostatic field (in N/C) at the origin (x=0)?
Add the two components...
Et=E1+E1=k4/20^2+k10^2/8^2
negative sign points to minus x axis, so in this case, E is to the right
To find the magnitude of the total electrostatic field at the origin, we need to calculate the individual electrostatic fields due to each particle at that point, and then add them together.
The formula for the electric field due to a point charge is given by:
E = k(q / r^2)
Where:
E is the electric field
k is the electrostatic constant (k = 9.0 x 10^9 N m²/C²)
q is the charge of the particle
r is the distance from the particle to the point where the electric field is being calculated.
Let's calculate the electric field due to the first particle (charge = +4.0 nC):
q1 = +4.0 nC
r1 = distance between the first particle and the origin = 0 - (-20.0) = 20.0 m
Using the formula, we get:
E1 = (9.0 x 10^9 N m²/C²) * (4.0 x 10^-9 C) / (20.0 m)²
Simplifying the expression, we get:
E1 = (9.0 x 4.0)/(20.0)² N/C
Now, let's calculate the electric field due to the second particle (charge = -10.0 nC):
q2 = -10.0 nC
r2 = distance between the second particle and the origin = 0 - 8.0 = -8.0 m (negative because it's in the opposite direction)
Using the formula, we get:
E2 = (9.0 x 10^9 N m²/C²) * (-10.0 x 10^-9 C) / (-8.0 m)²
Simplifying the expression, we get:
E2 = (-9.0 x 10) / 8.0² N/C
Finally, let's add the two electric fields together to get the total electric field at the origin:
E_total = E1 + E2
Substituting the values, we have:
E_total = (9.0 x 4.0 / 20.0²) + (-9.0 x 10 / 8.0²) N/C
Calculating this expression, we'll get the magnitude of the total electrostatic field at the origin.