A hollow sphere is charged to 14 µC. Find the potential (a) at its surface (b) inside the sphere (c) at a distance of 0.2m from the surface. The radius of the sphere is 0.3m. (Answer 42 ｘ 10⁴ V, 42 ｘ 10⁴ V, 25.2 ｘ 10⁴ V)  



Given:

Hollow sphere of charge = q = 14 ｘ 10⁻⁶ C  

Radius of sphere = r = 0.3 m

Distance = d = 0.2 m

∴ Coulomb's Constant = K = `\frac {1}{4ㄫε_0}` = 9 ｘ 10⁹ N m² C⁻²

To Find:

(a) Potential at the surface of sphere = `V_s` = ?

(b) Potential inside of sphere = `V_i` = ?

(c) Potential at distance d = `V_d` = ?


Solution:


The General formula for finding potential V is

V = k `\frac {q}{r}`


(a) Potential at the surface of sphere = `V_s` = ?

`V_s` = k `\frac {q}{r}`

putting values

`V_s` =  9 ｘ 10⁹ N m² C⁻² ｘ  `\frac {14 ｘ 10⁻⁶ C}{0.3 m}`

`V_s` =  9 ｘ 10⁹ N m² C⁻² ｘ  64.667 ｘ 10⁻⁶ C m⁻¹

`V_s` =  420 ｘ 10⁹⁻⁶ N m C⁻¹

`V_s` =  420 ｘ 10³ N m C⁻¹

or

`V_s` =  42 ｘ 10⁴ N m C⁻¹ ------------------Ans.



(b) Potential inside of sphere = `V_i` = ?

The charge distributes itself equally on the surface as well as inside the sphere, so

`V_s` =  `V_i` = 42 ｘ 10⁴ N m C⁻¹
  

(c) Potential at distance d = 0.2 m away from sphere:

Here the total distance r' = distance d + radius r

r' = 0.2 m + 0.3 m = 0.5 m

now

`V_s` = k `\frac {q}{r'}`

putting values

`V_d` =  9 ｘ 10⁹ N m² C⁻² ｘ  `\frac {14 ｘ 10⁻⁶ C
}{0.5 m}`

`V_d` =  9 ｘ 10⁹ N m² C⁻² ｘ  28 ｘ 10⁻⁶ C m⁻¹

`V_d` =  252 ｘ 10⁹⁻⁶ N m C⁻¹

`V_d` =  252 ｘ 10³ N m C⁻¹

or

`V_d` =  25.2 ｘ 10⁴ N m C⁻¹ ------------------Ans.



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.