You've got a conducting hollow sphere of rradius 0.2 m and a charge of 20 micro culumbs. What is the potential of the sphere at distance r from the center if:

r=0.1 m?
r=0.2 m?
r=0.3 m?

i need it step by step with explanations and solutions

To find the potential of a conducting hollow sphere at different distances from the center, you can use the formula for the potential due to a charged conducting sphere:

V = k * (Q / R)

Where:
- V is the potential
- k is the electrostatic constant (8.99 x 10^9 N m^2/C^2)
- Q is the charge of the sphere
- R is the radius of the sphere

Let's calculate the potential at each distance step by step:

1) For r = 0.1 m:
- substitute the given values into the formula:
V = (8.99 x 10^9 N m^2/C^2) * (20 x 10^-6 C) / 0.2 m
- perform the multiplication:
V = (8.99 x 10^9) * (20 x 10^-6) / 0.2
- simplify the expression:
V = 4.495 x 10^8 / 0.2
- divide:
V = 2.2475 x 10^9 N m^2/C

Therefore, the potential at r = 0.1 m is 2.2475 x 10^9 N m^2/C.

2) For r = 0.2 m:
- substitute the given values into the formula:
V = (8.99 x 10^9 N m^2/C^2) * (20 x 10^-6 C) / 0.2 m
- perform the multiplication:
V = (8.99 x 10^9) * (20 x 10^-6) / 0.2
- simplify the expression:
V = 4.495 x 10^8 / 0.2
- divide:
V = 2.2475 x 10^9 N m^2/C

Therefore, the potential at r = 0.2 m is also 2.2475 x 10^9 N m^2/C.

3) For r = 0.3 m:
- substitute the given values into the formula:
V = (8.99 x 10^9 N m^2/C^2) * (20 x 10^-6 C) / 0.3 m
- perform the multiplication:
V = (8.99 x 10^9) * (20 x 10^-6) / 0.3
- simplify the expression:
V = 4.495 x 10^8 / 0.3
- divide:
V = 1.4983 x 10^9 N m^2/C

Therefore, the potential at r = 0.3 m is 1.4983 x 10^9 N m^2/C.

In summary:
- The potential at r = 0.1 m is 2.2475 x 10^9 N m^2/C.
- The potential at r = 0.2 m is also 2.2475 x 10^9 N m^2/C.
- The potential at r = 0.3 m is 1.4983 x 10^9 N m^2/C.