Question

Two particles of charge ql = + 2 X 10- 9 C and q2 = +3 X 10- 9 C are placed 0.04 m from each other. What is the force of repulsion that each experiences?

Answer 1

Explanation:

this is the answer of the question

Answer 2

Answer:

the answer is

$3.375 \times {10}^{ - 5} N$

Explanation:

$\frac{1}{4\piε0} = 9 \times {10}^{9} N {m}^{2} {C}^{ - 2}$

r is the distance between two charges = 0.04m

q1,q2 are the charges.

Applying the Coloumb force of attraction or repulsion,

$F = \frac{1}{4\piε0} \times \frac{q1 \times q2}{ {r}^{2} } \\ = 9 \times {10}^{9} \times \frac{2 \times {10}^{ - 9} \times 3 \times {10}^{ - 9} }{ {(0.04)}^{2} } \\ = 9 \times {10}^{9} \times \frac{6 \times {10}^{ - 18} }{16 \times {10}^{ - 4} } \\ = 27 \times {10}^{ - 9} \times \frac{1}{8 \times {10}^{ - 4 } } \\ = 3.375 \times {10}^{( - 9 + 4)} \\ = 3.375 \times {10}^{ - 5} N$