Final answer:
The electric field magnitude is E/4 at 4m away from the charge, as the electric field strength varies inversely with the square of the distance according to Coulomb's Law. so, option A is the correct answer.
Step-by-step explanation:
The student's question pertains to the electric field produced by a point charge and how the electric field's magnitude changes with distance. According to Coulomb's Law and the electric field formula E = kQ/r², where E is the electric field, k is Coulomb's constant, Q is the charge, and r is the distance from the charge, the electric field varies inversely with the square of the distance from the charge.
Since at 2m the field is E, and we're looking for the distance where the field is E/4, we apply the inverse square law to determine that at double the distance (4m), the field will be one-fourth. Thus, the correct answer to the question is (a) 4m away.
In this question, we are given that an isolated point charged particle produces an electric field with magnitude E at a point 2m away from the charge. We are asked to determine the distance of a point where the field magnitude is E/4.
Since the electric field of a point charge is inversely proportional to the square of the distance, we can use the relationship E1/E2 = (r2/r1)^2, where E1 and E2 are the electric field magnitudes, and r1 and r2 are the distances.
Given that E1 = E and r1 = 2m, and we want to find r2 when E2 = E/4, we can solve for r2 as follows:
E1/E2 = (r2/r1)^2
E/(E/4) = (r2/2)^2
4 = (r2/2)^2
2 = r2/2
r2 = 4m
Therefore, the point where the electric field magnitude is E/4 is 4m away.