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

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

Step-by-step 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.