Two point charges; QA = +0.6 = −0.2 are separated by a distance of 5.0 cm. Find the following (#1-3) :
1. The electric force acting on each charge.
2. The electric force acting on each charge if their distance is doubled.
3. The electric force acting on each charge if charge is doubled.
Answers
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Here's your answer
Here it is given that,
q₁ = +0.6μC = +0.6 * 10⁻⁶ C
q₂ = -0.2μC = -0.2 * 10⁻⁶ C
r = 5cm = 0.05m
1)
F = kq₁q₂/r²
F = 9×10⁹ * 0.6 * 10⁻⁶ * 0.2 * 10⁻⁶/(0.05)²
F = 1.08 * 10⁹⁻⁶⁻⁶ / 0.0025
F = 432 * 10⁻³
F = 0.432 N
2)
If the distance is doubled, the force will reduce by 4 times
F = 0.432/4 = 0.108 N
3)
If both the charges are doubled, the force will increase by 4 times
F = 4 * 0.432
F = 1.728 N
Hope this is helpful to you
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Explanation:
since unit of charges is not given
=> if it is esu
=> q1 = +0.6 esu
q2 = -0.2 esu
r = 5cm
in CGS system 1/ 4π€o = 1
1:- F= 1× 0.6×(- 0.2) / 5²
= -12/25 ( - sign means force is attractive)
2:- distance is doubled
=> force = -12/ 100=> one fourth of initial
3:-if both charges are doubled
=> force acting is four times of it's initial value