Two point charges q₁= 10 µC and q₂= 5 µC are placed at a distance of 150 cm. What will be the Coulomb's force between them? Also, find the direction of the force. Ans. (0.2 N, the direction of repulsion)


Given Data:


1st point Charge = q₁ = 10 µC = 10 𝐱 10⁻⁶ C 

2nd point Charge = q₂ = 5 µC = 5 𝐱 10⁻⁶ C
 
Distance between the charges= r  = 150 cm = `\frac{150}{100}`m = 1.5 m

Constant of proportionality = K = 9 ｘ 10⁹ Nm²C⁻²


To Find:


The magnitude of the Coulomb's Force = F = ? 

The direction of the Coulomb's Force =  ? 



Solution:


According to Coulomb's Law

F = K `\frac{q₁q₂}{r²}`

by putting the values

F = 9 ｘ 10⁹ Nm²C⁻² `\frac{10 𝐱 10⁻⁶ C ｘ 5 ｘ 10⁻⁶ C}{(1.5 m)²}`

F = `\frac{9 ｘ 10 ｘ 5}{2.25}` ｘ 10⁹⁻⁶⁻⁶ N

by simplifying

F = 200 ｘ 10⁻³ N

or

F = 0.2 N  ----------------Ans.

The direction of Coulomb's is in the direction of repulsion.

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