Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
=
for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E = ![\frac{KQ}{r^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BKQ%7D%7Br%5E%7B2%7D%20%7D)
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = - ![\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes5.50%5Ctimes10%5E%7B-6%7D%20%20%7D%7B3.10%5E%7B2%7D%20%7D)
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E = ![\frac{KQ}{r^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BKQ%7D%7Br%5E%7B2%7D%20%7D)
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = - ![\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B9%5Ctimes10%5E%7B9%7D%5Ctimes5.50%5Ctimes10%5E%7B-6%7D%20%20%7D%7B8%5E%7B2%7D%20%7D)
E = - 0.77 x 10³ V/m