Answer:
e
Explanation:
From the given information:
Suppose Q = total charge of the sphere.
here, the electric field outside the sphere at distance R/2 can be expressed as:
[tex]E_1 = \dfrac{1}{4 \pi \varepsilon _o}* \dfrac{Q}{(R + \dfrac{R}{2})^2}[/tex]
where:
[tex]k = \dfrac{1}{4 \pi \varepsilon _o}[/tex]
[tex]E_1 = \dfrac{kQ}{(\dfrac{3R}{2})^2}[/tex]
[tex]E_1 = \dfrac{4kQ}{9R^2}[/tex]
For the electric field inside the sphere, we have:
[tex]E_2 = \dfrac{kQr}{R^3}[/tex]
here:
r = distance of the point from the center = R/2
R = radius of the sphere
∴
[tex]E_2 = \dfrac{kQ * \dfrac{R}{2}}{R^3}[/tex]
[tex]E_2 = \dfrac{kQ }{2R^2}[/tex]
As such, the ratio of the electric field outside the sphere to the one inside is:
[tex]\dfrac{E_1}{E_2} = \dfrac{ \dfrac{4kQ}{9R^2}}{ \dfrac{kQ }{2R^2}}[/tex]
[tex]\dfrac{E_1}{E_2} = \dfrac{4kQ}{9R^2} \times \dfrac{ 2R^2 }{kQ}[/tex]
[tex]\mathbf{\dfrac{E_1}{E_2} = \dfrac{8}{9}}[/tex]
For a solid uniformly charged sphere of radius R, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is - e) 8/9
field [tex]E2= \frac{k(Q)}{(3R/2)^2}[/tex]
= [tex]\frac{4}{9} \frac{k(Q)}{R^2}[/tex]
=
Thus, E2/E1
= (4/9)/(1/2)
= 8/9
Thus, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is - 8/9
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