Answer:
0 MN/C, 2.697 MV
4.1953 MN/C, 1.2586 MV
19.2642857143 MN/C, 2.697 MV
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
Electric field at r = 8 cm
E = 0 (inside)
Electric potential is given by
[tex]V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]
Electric potential is 2.697 MV
r = 30 cm
[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C[/tex]
Electric field is 4.1953 MN/C
[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV[/tex]
Electric potential is 1.2586 MV
r = R = 14 cm
[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C[/tex]
The electric field is 19.2642857143 MN/C
[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]
The potential is 2.697 MV
