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marissa [1.9K]
3 years ago
9

How much charge is required to raise an isolated metallic sphere of 1.0-meter radius to a potential of volts? Repeat for a spher

e 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?
Physics
1 answer:
frosja888 [35]3 years ago
3 0

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.

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