An electromagnetic generator consists of a coil that has 100 turns of wire, has an area of 400 cm2, and rotates at 60 rev/ s in
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Answer:
<em>3.15 N towards the positive x-axis</em>
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Explanation:
first charge has charge q1 = 10 μC = 10 x 10^-6 C
second charge has charge q2 = 20 μC = 20 x 10^-6 C
third charge has charge q3 = -30 μC = -30 x 20^-6 C
According to coulomb's law, force between two charged particle is given as
F =
Where
F is the force between the charges
k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.
Q is the magnitude of one charge
q is the magnitude of the other charge
is the distance between these two charges
For the force on q2 due to q1,
distance r between them = 0 - (-1.0) = 1 m
F = = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)
For the force on q2 due to q3,
distance between them = 2.0 - 0 = 2 m
F = = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)
Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]