Question

A charge q produces an electric field of strength 4E at a distance of d away. Determine the electric field strength at a distance of $\frac{1}{3}$d away.



a.) $\frac{1}{36}$E


b.) $\frac{1}{9}$E


c.) 36E


d.) 9E

Answer 1

Answer:

c.) 36E

Explanation:

The magnitude of the electric field is given by the expression

$E=k \frac{q}{d^2}$ (1)

where k is the Coulomb's constant, q is the charge that generates the field, and d is the distance from the charge.

In this problem, we have that the magnitude of the field at a distance d is 4E, so we can rewrite the previous equation as

$4E = k\frac{q}{d^2}$

Now we want to determine the electric field at a distance of $d'=\frac{1}{3}d$ away. Substituting into (1), we find

$E' = k \frac{q}{d'^2}=k \frac{q}{(\frac{1}{3}d)^2}=9 k \frac{q}{d^2}$ (2)

We also know that

$4E = k\frac{q}{d^2}$ (3)

So combining (2) with (3), we find a relationship between the original field and the new field:

$E' = 9 \cdot (4E) = 36E$

Answer 2

Answer:

36E

Explanation: