Answer:
[tex]q_1 = 1,112 * 10 ^ {-16}[/tex]
Explanation:
A) First we will perform the exercise for the 1m radius sphere.
The key is to understand the electric potential of the sphere as well
[tex]r_1 = 1m[/tex]
[tex]r_2 = 0.1m[/tex]
[tex]V_p = 1 * 10 ^ 6V[/tex]
Understanding that,
[tex]V_p=\frac{q} {4\pi\epsilon*r}[/tex]
Where,
[tex]\epsilon = \epsilon_0 \epsilon_r[/tex]
[tex]\epsilon = 8.85 * 10^{-12}[/tex]
Clearing q of the electric potential equation,
[tex]q = V_p (4 \pi) (\epsilon) (r)[/tex]
So,
[tex]q_1 = (1 * 10 ^ -6) (4 \pi) (8.85 * 10 ^{ - 12}) (1)[/tex]
[tex]q_1 = 1,112 * 10 ^ {- 16} c[/tex]
While for the second sphere
[tex]q_2 = (1 * 10 ^{ -6}) (4 \pi) (8.85 * 10^ {- 12}) (0.1)[/tex]
[tex]q_2 = 1,112 * 10 ^ {- 17}} c[/tex]
B) It is not entirely true, by obtaining a difference of one tenth, it is possible to understand that the larger the diameter, the greater the voltage reached.
