Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
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<span>a=Δω/Δt
</span><span>a=2π*Δf/Δt
</span><span>a=2π*(f2-f1)/Δt
</span>
<span>f1=f2-a*Δt/2π
</span><span>f2=800/60 rev/sec
</span><span>a=-42 rad/sec^2
</span><span>Δt=1.75sec
</span><span>so
f1=25 rev/sec
f1=1500 rev/min</span>
Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
<h3>What is electrical force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The given data in the problem is
q₁ is the negative charge = 6 µC=6×10⁻⁶ C
q₂ is the positive charge = 3 µC=3×10⁻⁶ C
r is the distance between the charges=0.002 m
is the electric force =?
The value of electric force will be;
Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
To learn more about the electrical force refer to the link;
brainly.com/question/1076352
Answer:
WHICH CHALENGE >>>>>>>>>>>>>>>>>>>>>>
Explanation:
