three equal charges of 10×10^-8 c respectively , each located at the corners of a right triangle whose sides are 15cm,20cm,25cm respectively. find the force exerted on the charge located at the 90° angle
Answers
force exerted on the charge located at the 90° is 4.6 × 10^-3 N (approx).
let ABC is a triangle where AB = 15cm, BC = 20cm and CA = 25cm
AB² + BC² = 15² + 20² = 625 = CA² = 25²
from Pythagoras, CA is hypotenuse so, B is right angle.
so, we have to find out force exerted on B due to A and C.
Force due to A on B, F1 = K(10 × 10^-8)²/(AB)²
= (9 × 10^9 × 10^-14)/(15 × 10^-2)²
= (9 × 10^-5)/(0.0225)
= 400 × 10^-5 N
= 4 × 10^-3 N
similarly, force due to C on B , F2 = K(10 × 10^-8)²/(CB)
= (9 × 10^9 × 10^-14)/(20 × 10^-2)²
= (9 × 10^-5)/(0.04)
= 2.25 × 10^-3 N
as angle between F1 and F2 is 90°
so, resultant force , F = √(F1² + F2²)²
= √(4² + 2.25²) × 10^-3
= 4.5893 × 10^-3 ≈ 4.6 × 10^-3 N
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let ABC is a triangle where AB = 15cm, BC = 20cm and CA = 25cm
AB² + BC² = 15² + 20² = 625 = CA² = 25²
from Pythagoras, CA is hypotenuse so, B is right angle.
so, we have to find out force exerted on B due to A and C.
Force due to A on B, F1 = K(10 × 10^-8)²/(AB)²
= (9 × 10^9 × 10^-14)/(15 × 10^-2)²
= (9 × 10^-5)/(0.0225)
= 400 × 10^-5 N
= 4 × 10^-3 N
similarly, force due to C on B , F2 = K(10 × 10^-8)²/(CB)
= (9 × 10^9 × 10^-14)/(20 × 10^-2)²
= (9 × 10^-5)/(0.04)
= 2.25 × 10^-3 N
as angle between F1 and F2 is 90°
so, resultant force , F = √(F1² + F2²)²
= √(4² + 2.25²) × 10^-3
= 4.5893 × 10^-3 ≈ 4.6 × 10^-3 N