Question

A sphere has a surface uniformly charged with 3.30 C. At what distance from its center is the potential 4.50 MV

Answer 1

Answer:

r = 6.6 x 10³ m = 6600 m

Explanation:

The potential at a distance from a charged sphere can be given as follows:

$V = \frac{kq}{r}\\\\r = \frac{kq}{V}$

where,

r = distance = ?

k = Colomb Constant = 9 x 10⁹ Nm²/C²

q = charge on sphere = 3.3 C

V = potential = 4.5 MV = 4.5 x 10⁶ V

Therefore,

$r = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.3\ C)}{4.5\ x\ 10^6\ V}$

r = 6.6 x 10³ m = 6600 m