Question

(a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?

Answer 1

Answer:

Part a)

$V = 7.2 \times 10^{11} Volts$

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.

Explanation:

Part a)

As we know that the potential near the surface of metal sphere is given by the equation

$V = \frac{kQ}{R}$

here we have

Q = 8 C

R = 10.0 cm

now we have

$V = \frac{(9\times 10^9)(8 C)}{0.10}$

$V = 7.2 \times 10^{11} Volts$

Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

here we can assume the sphere is placed at vacuum so that there is no break down of air.