Question

calculate the force of attraction between an electron and a body with charge +1.0cand it placed at a distance of 1.5m a part k=9.0*10^19 Nm^2/c^2​

Answer 1

6.4 × 10^-10 N is the required answer.

GIVEN:

• Charge of electron, e = 1.6 × 10^-19 C
• Charge of the body, q = 1 C
• Distance of seperation, r = 1.5 m
• k = 9 × 10¹⁹ Nm²/C²

TO FIND:

• Electrostatic force of attraction, F

FORMULA:

According to coulomb's law of electrostatics,

F = kqe/r²

SOLUTION:

$F = \dfrac{kqe}{ {r}^{2} } \\ \\ F = \dfrac{9 \times {10}^{9} \times 1 \times 1.6 \times {10}^{ - 19} }{ {(1.5)}^{2} } \\ \\ F = \dfrac{9 \times 1.6 \times {10}^{9} \times {10}^{ - 19} }{1.5 \times 1.5} \\ \\ F = \dfrac{9 \times 1.6 \times {10}^{(9 - 19)} }{1.5 \times 1.5} \\ \\ F = \dfrac{9}{1.5} \times \dfrac{1.6}{1.5} \times {10}^{ - 10} \\ \\ F = \dfrac{3}{0.5} \times \dfrac{1.6}{1.5} \times {10}^{ - 10} \\ \\ F = \dfrac{3 \times 1.6}{0.5 \times 1.5} \times {10}^{ - 10} \\ \\ F = \dfrac{1 \times 1.6}{0.5 \times 0.5} \times {10}^{ - 10} \\ \\ F = \dfrac{1.6}{0.25} \times {10}^{ - 10} \\ \\ F = 6.4 \times {10}^{ - 10} \: newton$

ANSWER:

The electrostatic force of attraction acting between the electron and the given charge is 6.4 × 10^-10 N.

COULOMB'S LAW OF ELECTROSTATICS:

• It is also called coulomb's inverse square law.

• The law states that, the electrostatic force of attraction between the two charges seperated by a distance r is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

• The force is attractive when there are unlike charge as in the question.

• The force is repulsive when like charges are present.