If the original electric field is 1.5x10^-4 N/C and the distance is reduced to one-half the original distance, what is the magnitude of the new electric field?

To find the magnitude of the new electric field, we need to understand how the electric field depends on distance. The electric field due to a point charge is inversely proportional to the square of the distance from the charge. This relationship can be expressed mathematically as:

E ∝ 1/r^2

Where E is the electric field and r is the distance from the charge.

Now, in this problem, the distance is reduced to one-half of the original distance. Let's call the original distance "d", and the new distance "d/2". We need to find the new electric field when the distance is reduced to one-half.

Using the inverse square relationship, we can write the following:

E1/E2 = (r2/r1)^2

Where E1 is the original electric field, E2 is the new electric field, r1 is the original distance, and r2 is the new distance.

Substituting the given values:

E1/E2 = (d/2)^2 / d^2

Simplifying:

E1/E2 = 1/4

To find E2, we can rearrange the equation:

E2 = E1 * (1/E1/E2)

E2 = E1 * 4

Now, substitute the value of E1, which is 1.5x10^-4 N/C, into the equation:

E2 = 1.5x10^-4 N/C * 4

E2 = 6x10^-4 N/C

Therefore, the magnitude of the new electric field is 6x10^-4 N/C.