Answer:
The point that would have the same electric field as P is [tex]z = 0.108 \ m[/tex] from the center of the sphere.
Explanation:
From the question we are told that
The radius of the sphere is [tex]r = 0.30 \ m[/tex]
The Electric field at point P is [tex]E = 15000N/C[/tex]
The distance of point P from the center is [tex]D = 0.50 \ m[/tex]
Since the electric is directed radially outward it mean this it would be felt both inside and outside the sphere
The Electric field inside the sphere at a distance z is mathematically represented as
[tex]E_i = \frac{k q x}{r^3}[/tex]
where k is the coulomb's constant with a value [tex]9 *10^9 \ kg \cdot m^3 \cdot s^{-4 } \cdort A^{-2 }[/tex]
q is the charge
The Electric field inside the sphere at a distance D is mathematically represented as
[tex]E _o = \frac{k q}{D^2}[/tex]
To obtain the point of equal electric field
[tex]E_i = E_o[/tex]
[tex]\frac{k q z}{r^3} = \frac{kq }{D^2}[/tex]
We have that
[tex]z = \frac{r^3 }{D^2}[/tex]
Substituting values
[tex]z = \frac{(0.3)^3 }{(0.5)^2}[/tex]
[tex]z = 0.108 \ m[/tex]
