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 = 14ㄫε0 = 9 ｘ 10⁹ N m² C⁻²

To Find:

(a) Potential at the surface of sphere = Vs = ?

(b) Potential inside of sphere = Vi = ?

(c) Potential at distance d = Vd = ?


Solution:


The General formula for finding potential V is

V = k qr


(a) Potential at the surface of sphere = Vs = ?

Vs = k qr

putting values

Vs =  9 ｘ 10⁹ N m² C⁻² ｘ  14ｘ10⁻⁶C0.3m

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

Vs =  420 ｘ 10⁹⁻⁶ N m C⁻¹

Vs =  420 ｘ 10³ N m C⁻¹

or

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



(b) Potential inside of sphere = Vi = ?

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

Vs =  Vi = 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

Vs = k qr′

putting values

Vd =  9 ｘ 10⁹ N m² C⁻² ｘ  14ｘ10⁻⁶C0.5m

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

Vd =  252 ｘ 10⁹⁻⁶ N m C⁻¹

Vd =  252 ｘ 10³ N m C⁻¹

or

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



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