Question

How much charge is required to raise an isolated metallic sphere of 1.0-meter radius to a potential of volts? Repeat for a sphere of 1.0-cm radius. (b) Why use a large sphere in an electrostatic generator, since the same potential can be achieved for a smaller charge with a small sphere?

Answer 1

Answer:

$q_1 = 1,112 * 10 ^ {-16}$

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

$r_1 = 1m$

$r_2 = 0.1m$

$V_p = 1 * 10 ^ 6V$

Understanding that,

$V_p=\frac{q} {4\pi\epsilon*r}$

Where,

$\epsilon = \epsilon_0 \epsilon_r$

$\epsilon = 8.85 * 10^{-12}$

Clearing q of the electric potential equation,

$q = V_p (4 \pi) (\epsilon) (r)$

So,

$q_1 = (1 * 10 ^ -6) (4 \pi) (8.85 * 10 ^{ - 12}) (1)$

$q_1 = 1,112 * 10 ^ {- 16} c$

While for the second sphere

$q_2 = (1 * 10 ^{ -6}) (4 \pi) (8.85 * 10^ {- 12}) (0.1)$

$q_2 = 1,112 * 10 ^ {- 17}} c$

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.