Numerical Problem No. 13.3 (Electrostatics) Physics SSC - II (Class 10)

The force of repulsion between two identical positive charges is 0.8 N, when the charges are 0.1 m apart. Find the value of each charge.  Ans. (9.4 𝐱 10⁻⁷ C)

Given Data:

Coulomb's Force = F = 0.8 N

Distance between the charges= r  = 0.1 m

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

To Find:

Magnitude (value) of each charge = q = ? 

Solution:

According to Coulomb's Law

F = K `\frac{q₁q₂}{r²}`

but we were given identical charges ie. q₁ = q₂ = q hence

F = K `\frac{q ｘ q}{r²}`

F = K `\frac{q²}{r²}`

or

q² =  `\frac{Fr²}{K}`

by putting the values

q² = `\frac{0.8 N ｘ (0.1 m)²}{9 ｘ 10⁹ Nm²C⁻² }`

q² = `\frac{0.8 N ｘ 0.01 m²}{9 ｘ 10⁹ Nm²C⁻² }`


q² = `\frac{0.008 N m²}{9 ｘ 10⁹ Nm²C⁻² }`

{ Nm² will cancel from the numerator and denominator}

q² = 8.889 ｘ 10⁴ｘ 10⁻⁹ C²

q² = 8.889ｘ 10⁻¹³ C²

or

q² = 88.889ｘ 10⁻¹⁴ C²

by taking square root on both sides we get

q = 9.428 ｘ 10⁻⁷ C

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