Two charges repel each other with a force of 0.1 N when they are 5 cm apart. Find the forces between the same charges when they are 2 cm apart. Ans. (0.62 N)


Given:

Part-1
Coulomb's Force = F = 0.1 N
Distance between the charges= r  = 5 cm = `\frac{5}{100}` = 0.05 m
Constant of proportionality = K = 9 ｘ 10⁹ Nm²C⁻²
Part-2
Distance between the charges= r  = 2 cm = 0.02 m

To Find:

The magnitude of the Coulomb's Force between the charges when they are (2cm = 0.02m) apart = F = ?

Solution: 

To find the Coulomb's force we also need to find first the magnitude of charges from the data of part-1 so,
According to Coulomb's Law

F = K `\frac{q₁q₂}{r²}`

or

q₁q₂ = `\frac{Fr²}{K}`

by putting the values

q₁q₂ = `\frac{0.1 N ｘ (0.05)²}{9 ｘ 10⁹ Nm²C⁻² }`

q₁q₂ = `\frac{0.1 ｘ 0.0025}{9}`ｘ 10⁻⁹ C²

q₁q₂ = `\frac{0.0025}{9}`ｘ 10⁻⁹ C²

q₁q₂ = `\frac{25}{9}` 10⁻⁴ｘ 10⁻⁹ C²

q₁q₂ = 2.7777ｘ 10 ⁻¹³ C²

for the same charges and the data of part 2, we are again using Coulomb's law equation

F =  K `\frac{q₁q₂}{r²}`

by putting values

F = 9 ｘ 10⁹ Nm²C⁻² `\frac{ 2.7777ｘ 10⁻¹² C²}{(0.02)²}`

F = `\frac{9 ｘ 2.777}{0.004}` 10⁹ ï½˜ 10⁻¹³ N

F = `\frac{25 ｘ 1000}{4}` ｘ 10⁻⁴ N

F = `\frac{32.4}{4}` ｘ 1000ｘ 10⁻⁴ N

F = 6.25ｘ 10³ï½˜ 10⁻⁴ N

F = 6.25ｘ 10⁻¹ N

F = 0.625 N


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