Question

The electrostatic force of repulsion between two electrons at a distance 1m is

Answer 1

It is given that two electrons are separated by distance of 1m We are required to find the electrostatic force.

It is given by formula

$\mathcal{ \: F = \dfrac{1}{4 \pi\ \: \mathcal{E_0} } } \: \: \dfrac{q_1 \: q_2}{r {}^{2} }$

where,

• F denotes electrostatic force
• 1/4πε₀ = 9×10⁹Nm²/C²
• q₁ ,q₂ denotes charges
• r denotes distance seperated

Charge of electron is given by -1.6×10⁻¹⁹C

Substituting the values,

$\mathcal{ \: F = 9 \times 10 {}^{9} \times \dfrac{ - 1.6 \times 10 {}^{ - 19} \times - 1.6 \times 10 {}^{ - 19} }{ 1 {}^{2} } }$

$\mathcal{ \: F = 9 \times 10 {}^{9} \times { - 1.6 \times 10 {}^{ - 19} \times - 1.6 \times 10 {}^{ - 19}}}$

$\mathcal{ \: F = 9 \times 10 {}^{9 - 19 - 19} \times 1.6 \times 1.6 }$

$\mathcal{ \: F = 9 \times 10 {}^{ - 29} \times 2.56 }$

$\mathcal{ \: F =23.56 \times 10 {}^{ - 29}N }$

Know more ::

$\mathcal{ \: F = \dfrac{1}{4 \pi\ \: \mathcal{E_0} } } \: \: \dfrac{q_1 \: q_2}{r {}^{2} }$

• Directly proportional to magnitude of charges q₁ ,q₂
• Inversly proportional to square of seperation r²

• ε₀= 8.85 × 10⁻¹² C²/Nm² which is called permitivity of free space
• 1/4πε₀ = 9×10⁹Nm²/C² which is constant (k)