When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
1 answer:
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Answer:
ac = 2.86 m / s²
Explanation:
Image can detail the system to determine the force in the FA to understand the system into the applicated force
m = 100 kg , L = 3 m
∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0
Ay = 768.86 N
∑ Mₐ = α * I ₐ
I ₐ = m * L² / 3 ⇒ I ₐ = 100 kg * 4² m / 3
Replacing
P * sin (45) * 3 = α * 100 kg * 4² m / 3
α = 1.193 rad / s²
ac = α *2 ⇒ ac = 1.193 rad / s² * 2
ac = 2.86 m / s²
The charges have opposite sign and magnitude
Explanation:
The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
In this problem, we have:
F = 3.60 N is the force between the two charges
r = 30 cm = 0.30 m is their separation
The two charges have same magnitude, so
So we can rewrite the equation as
And solving for q:
Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so
Learn more about electric force:
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