Question

what is the magnitude of the electric field produced by a charge of magnitude 8.30 μC at a distance of (a) 1.00 mm and (b) 3.00 mm?

Answer 1

The magnitude of the electric field produced by a charge of 8.30 µC at a distance of 1.00 mm is 7.49 x 10^10 N/C, and at a distance of 3.00 mm is 8.32 x 10^9 N/C, calculated using Coulomb's law for electric field.

Step-by-step explanation:

To calculate the magnitude of the electric field produced by a charge, we use Coulomb's Law for the electric field E, which is given by the equation E = k * |q| / r^2, where:

• k is Coulomb's constant (8.99 x 10^9 N m^2/C^2),
• q is the charge (in Coulombs), and
• r is the distance from the charge (in meters).

For a charge of 8.30 μC (which is 8.30 x 10^-6 C), the electric field at distances of:

1. 1.00 mm (or 0.001 m) would be E = (8.99 x 10^9) * (8.30 x 10^-6) / (0.001^2) = 7.49 x 10^10 N/C,
2. 3.00 mm (or 0.003 m) would be E = (8.99 x 10^9) * (8.30 x 10^-6) / (0.003^2) = 8.32 x 10^9 N/C.

Therefore, the magnitudes of the electric field at 1.00 mm and 3.00 mm from the charge are 7.49 x 10^10 N/C and 8.32 x 10^9 N/C, respectively.