Question

Electric force of repulsion between two electrons at a distance of 1 metre is?

Answer 1

F=kq²/r²

q=1.6×10^-19

r=1

and k=9×10^9...

hey i dont have calculator so i am not calculating ....

Answer 2

Given:

• 2 electrons are separated by a distance of 1 metre.

To find:

• Force of repulsion ?

Calculation:

As per Coulomb's Law of Electrostatic Force, we can say:

$\boxed{F = \dfrac{k(q1 \times q2)}{ {d}^{2} } }$

• 'k' is Coulomb's Constant , q1 and q2 represent the charges and 'd' is separation distance.

• Value of Coulomb's Constant is 9 × 10⁹.

$\implies F = \dfrac{9 \times {10}^{9} \times (e\times e)}{ {(1)}^{2} }$

$\implies F = \dfrac{9 \times {10}^{9} \times {e}^{2} }{ {(1)}^{2} }$

• Charge of electron = 1.6 × 10^(-19) C.

$\implies F = \dfrac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19}) }^{2} }{ {(1)}^{2} }$

$\implies F = 9 \times {10}^{9} \times 2.56 \times {10}^{ - 38}$

$\implies F = 23.04 \times {10}^{ - 29} \: N$

So, net force is 23.04 × 10^(-29) N