The electric field is C. Four times as strong
Explanation:
The magnitude of the electric field around a test charge is given by
[tex]E=k\frac{q}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
We notice that the magnitude of the field is inversely proportional to the square of the distance:
[tex]E\propto \frac{1}{r^2}[/tex]
In this problem, we are comparing the field at a distance of
r = 2 mm
with the field at a distance of
r' = 4 mm
So the distance has doubled:
r' = 2r
This means that the new field will be
[tex]E\propto \frac{1}{r'^2}=\frac{1}{(2r)^2}=\frac{1}{4}(\frac{1}{r^2})[/tex]
This means that the field at 4 mm is one fourth as strong as the field at 2 mm: therefore, the field at 2 mm is four times as strong as the field at 4 mm.
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