what is the magnitude and direction of the electric field 20.0 cm cm directly above an isolated 33.0 10 c charge?
To find the magnitude and direction of the electric field directly above a charge, we can use Coulomb's Law. Coulomb's Law states that the electric field, E, produced by a point charge, q, at a distance, r, from the charge is given by the equation:
E = k * (q / r^2)
Where:
- E is the magnitude of the electric field
- k is the Coulomb's constant (9.0 x 10^9 N m^2 / C^2)
- q is the charge
- r is the distance from the charge
In this case, we have:
- Charge, q = -33.0 x 10^-6 C (negative because it is an isolated charge)
- Distance, r = 20.0 cm = 0.20 m (converted to meters)
Now, let's plug these values into the equation to find the magnitude of the electric field:
E = (9.0 x 10^9 N m^2 / C^2) * (-33.0 x 10^-6 C) / (0.20 m)^2
Calculating this expression will give us the magnitude of the electric field.
To find the magnitude and direction of the electric field above a point charge, you can use Coulomb's Law. Coulomb's Law states that the electric field at a distance r from a point charge Q is given by the equation:
E = k * (Q / r^2)
Where:
E is the electric field
k is the electrostatic constant (k = 9.0 x 10^9 N m^2/C^2)
Q is the charge
r is the distance from the charge
In this case, the charge Q is given as -33.0 x 10^-6 C (negative because it's a negative charge). The distance r is given as 20.0 cm (which we'll convert to meters).
First, let's convert the distance from centimeters to meters:
20.0 cm = 0.20 m
Now we can calculate the magnitude of the electric field:
E = (9.0 x 10^9 N m^2/C^2) * (33.0 x 10^-6 C) / (0.20 m)^2
E = (9.0 x 10^9 N m^2/C^2) * (33.0 x 10^-6 C) / (0.04 m^2)
E = (9.0 x 10^9 N m^2/C^2) * (33.0 x 10^-6 C) / (0.0016 m^2)
Calculating this expression gives a magnitude of the electric field, E, in Newtons per Coulomb (N/C).
Now, to find the direction of the electric field, we need to consider the fact that the field lines radiate outward from a positive charge and inward toward a negative charge. In this case, we have a negative charge (-33.0 x 10^-6 C), so the electric field lines would point inward, towards the charge.
Therefore, the magnitude of the electric field directly above the isolated -33.0 x 10^-6 C charge is calculated in Step 5 as E N/C, and the direction is inward, towards the charge.