At a distance of 2.0 m. from a +8.0 x 10^-6 C charge, the magnitude and direction of the electric field will be?
Coulomb's law:
F = k q1 q2/r^2
E = force on unit charge
E = F/1 = k q /r^2
= 9*10^9 * 8*10^-6 /4
1.8 *10^4 Newtons/Coulomb
like charges repel so in direction away from positive charge
To find the magnitude and direction of the electric field at a distance of 2.0 m from a charge of +8.0 x 10^-6 C, you can use Coulomb's law. Coulomb's law states that the electric field at a point around a charged object is directly proportional to the charge and inversely proportional to the square of the distance from the charge.
The formula for Coulomb's law is:
E = k * (q / r^2)
Where:
E is the electric field strength,
k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2),
q is the charge, and
r is the distance between the charge and the point where the electric field is measured.
Given:
q = +8.0 x 10^-6 C (positive charge)
r = 2.0 m
Now, substituting the values into the formula, we get:
E = (8.99 x 10^9 Nm^2/C^2) * (+8.0 x 10^-6 C) / (2.0 m)^2
Calculating this expression, we find:
E = 8.99 x 10^9 Nm^2/C^2 * (8.0 x 10^-6 C) / (2.0 m * 2.0 m)
E = 1.7976 x 10^3 N/C
So, the magnitude of the electric field is 1.7976 x 10^3 N/C.
Since the charge is positive, the electric field will point away from the charge, radially outward.
Therefore, the magnitude of the electric field at a distance of 2.0 m from the +8.0 x 10^-6 C charge is 1.7976 x 10^3 N/C, directed radially outward.