The electric field changes by a factor 4/9
Explanation:
The electric field at a point outside a hollow charged sphere is identical to that produced by a single point charge Q, therefore its strength is given by
[tex]E=k\frac{Q}{r^2}[/tex]
where
k is the Coulomb constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
In this problem, we have a charge Q accumulated on the dome of the generator of radius R. The electric field strength at a distance of 2R from the sphere centre is
[tex]E=\frac{kQ}{(2R)^2}=\frac{1}{2}\frac{kQ}{R^2}[/tex]
Then, the probe is moved to a new distance of
r = 3R
From the centre of the sphere: therefore, the new electric field will be
[tex]E'=\frac{kQ}{(3R)^2}=\frac{1}{9}\frac{kQ}{R^2}[/tex]
So, the electric field has changed by a factor
[tex]\frac{E'}{E}=\frac{1/9}{1/4}=\frac{4}{9}[/tex]
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