Question

The coulomb's force b/w the 2 point charges 10 muC & 5 muC placed at a distance of 150cm is ​

Answer 1

Answer:

The coulomb's force b/w the 2 point charges 10 muC & 5 muC placed at a distance of 150cm is .2 N.

Explanation:

Coulomb's force formula is

$F = k \frac{q1 * q2}{r^{2} }$

Where

F = Force acting between two point of charges.

k = Constant proportionality = $8.998 * 10^{9} \frac{N.m^{2} }{C^{2} }$

q1 = Charge of one point charge = $10$μC = $10 * 10^{-6}$ C

q2 = Charge of another one point charge = $5$ μC = $15 * 15^{-6}$ C

r = distance between the two point charges = 150 cm = 1.5 meter.

The coulomb's force b/w the 2 point charges 10 muC & 5 muC placed at a distance of 150cm is

$8.998 * 10^{-9} * \frac{10*10^{-6} * 5 * 10^{-6} }{1.5^{2} }$   $N$

$= .2 N$

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