Answer:
4. At twice the distance the field strength will be 1/4 of the original value
Explanation:
The magnitude of the electric field produced by an isolated poitn charge is
[tex]E=k\frac{q}{r^2}[/tex]
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, the distance from the charge is doubled from 4 m to 8 m:
[tex]r'=2 r[/tex]
Therefore, the new electric field will be
[tex]E'=k\frac{q}{(r')^2}=k\frac{q}{(2r)^2}=k\frac{q}{4r^2}=\frac{1}{4}E[/tex]
So, the field strength is 1/4 of the original value.
By doubling the distance between two charges the field strength will be [tex]\dfrac{1}{4} th[/tex] of the original value.
The formula for the magnitude of the electric field is given by
[tex]E=k\dfrac{q}{r^2}[/tex]
Here
k is the Coulomb's constant
q is the charge
r is the distance from the charge
It is given in the problem that the distance from the charge is doubled from 4 m to 8 m:
[tex]r'=2r[/tex]
So to find the new electric field
[tex]E'=k\dfrac{q}{r^2} = k\dfrac{q}{(2r)^2} =\dfrac{1}{4} k\dfrac{q}{r^2} =\dfrac{1}{4} E[/tex]
Thus By doubling the distance between two charges the field strength will be [tex]\dfrac{1}{4} th[/tex] of the original value.
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