what is the magnitude of the electric field produced by a charge of magnitude 6.00 micro coulombs at a distance of a. 1.00m b. 2.00m
E=kq/distance
To calculate the magnitude of the electric field produced by a charge, you can use Coulomb's Law. Coulomb's Law states that the magnitude of the electric field (E) created by a point charge is given by the equation:
E = k * Q / r^2
Where:
- E is the electric field
- k is the electrostatic constant, approximately 9.0 x 10^9 N·m^2/C^2
- Q is the magnitude of the charge
- r is the distance from the charge (in meters)
For part (a) when the distance is 1.00m, and the magnitude of the charge is 6.00 microcoulombs (6.00 × 10^-6 C), we can substitute these values into the equation:
E = (9.0 × 10^9 N·m^2/C^2) * (6.00 × 10^-6 C) / (1.00 m)^2
Evaluating the equation gives:
E = (9.0 × 10^9 N·m^2/C^2) * (6.00 × 10^-6 C) / (1.00 m)^2
= (9.0 × 10^9 N·m^2/C^2) * (6.00 × 10^-6 C) / 1.00 m^2
= 54 N/C
Therefore, the magnitude of the electric field at a distance of 1.00m is 54 N/C.
For part (b) when the distance is 2.00m, we can use the same equation and substitute the new distance value:
E = (9.0 × 10^9 N·m^2/C^2) * (6.00 × 10^-6 C) / (2.00 m)^2
= (9.0 × 10^9 N·m^2/C^2) * (6.00 × 10^-6 C) / 4.00 m^2
= 27 N/C
Therefore, the magnitude of the electric field at a distance of 2.00m is 27 N/C.