Question

A charged particle creates an electric field of magnitude 300 N/C at a point 0.800 m away. What is the difference in the field magnitude between that point and one at 0.400 m ?

Answer 1

The difference in the electric field magnitude between the point at 0.800 m and the point at 0.400 m from a charged particle is 900 N/C, given that the field magnitude at 0.800 m is 300 N/C.

Step-by-step explanation:

The question involves the calculation of the electric field differences at two points located at different distances from a charged particle. The electric field produced by a point charge diminishes with the square of the distance from the charge, according to Coulomb's law. Given that the electric field is 300 N/C at a distance of 0.800 m, we can use the formula for the electric field E = k * q / r², where k is Coulomb's constant, q is the charge, and r is the distance from the charge, to find the electric field at the closer distance of 0.400 m. By calculating the electric field at 0.400 m and subtracting the original electric field at 0.800 m, we can find the difference in field magnitudes between the two points. The electric field at 0.400 m will be four times stronger than at 0.800 m because the distance is halved (and squaring the inverse of the distance will result in a factor of 4).

Here is the calculation step-by-step:

Therefore, the electric field difference between the two points would be 300 N/C * 4 - 300 N/C = 1200 N/C - 300 N/C = 900 N/C.