Two positive point charges of 15 ｘ 10⁻¹⁰ C and 13 ｘ10⁻¹⁰ C are placed 12 cm apart. Find the work done in bringing the two charges 4 cm closer. (Answer 7.31 ｘ 10⁻⁸ J)  



Given:

Let 

1st positive point charge = q₁ = 15 ｘ 10⁻¹⁰ C 

2nd positive point charge = q₂ = 13 ｘ 10⁻¹⁰ C  

Separation distance = r = 12 cm = 0.12 m

Separation distance after bringing the charge 4 cm  closer = r' =  r - 4 cm = 12 cm -4 cm = 8cm = 0.08 m

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

To Find:

Work done  = W = ?


Solution:

The diagram can be drawn according to the given conditions are  as below



The work done in moving the charge q₂ 4cm closer can be calculated as 

W = kq₁q₂`\{1/{r'} - 1/r}`

by putting the corresponding values

W = 9ｘ10⁹ Nm²C⁻² ｘ 15ｘ10⁻¹⁰ C ｘ15ｘ10⁻¹⁰ C ｘ `\{1/{0.08 m} - 1/0.12 m}`

W = 1.755 ｘ 10⁻⁸ N m² ｘ `\frac {0.12 m - 0.08 m }{(0.08 m)(0.12 m)}`

W = 1.755 ｘ 10⁻⁸ N m² ｘ `\frac {0.04 m }{0.0096 m^2}`

W = 1.755 ｘ 10⁻⁸ N m² ｘ 4.167 m⁻¹

W = 7.31 ｘ 10⁻⁸ N m

W = 7.31 ｘ 10⁻⁸ J ----------Ans.


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