Answer:
(a) 0 N/C
(b) - 2.7 x 10^6 N/C
(c) - 1.6 x 10^6 N/C
Explanation:
q = -12 micro Coulomb
(a) According to the Gauss's theorem
[tex]\int E.ds=\frac{q_{enclosed}}{\epsilon _{0}}[/tex]
Here, E be the electric field, ds be the area of gaussian surface and q be the charge enclosed.
In this case, the charge enclosed is zero, so the electric field is zero.
E = 0
(b) r = 20 cm = 0.2 m
Use the formula for the electric field
[tex]E = \frac{1}{4\pi \epsilon _{0}}\times \frac{q}{r^{2}}[/tex]
[tex]E = -\frac{9\times 10^{9}\times 12\times 10^{-6}}{0.2\times 0.2}[/tex]
E = - 2.7 x 10^6 N/C
(c) r = 20 + 6 = 26 cm = 0.26 m
Use the formula for the electric field
[tex]E = \frac{1}{4\pi \epsilon _{0}}\times \frac{q}{r^{2}}[/tex]
[tex]E = -\frac{9\times 10^{9}\times 12\times 10^{-6}}{0.26\times 0.26}[/tex]
E = - 1.6 x 10^6 N/C
