A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is
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Answer:
x = 41.2 m
Explanation:
The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.
In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract
F = F₁₃ - F₂₃
let's write the expression for each force, let's set a reference frame on the charge q1
F₁₃ =
F₂₃ =
they ask us that the net force be zero
F = 0
0 = F₁₃ - F₂₃
F₁₃ = F₂₃
k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}
q1 / x2 = q2 / (d-x) 2
(d-x)² = x²
we substitute
(100 - x)² = 2/1 x²
100- x = √2 x
100 = 2.41 x
x = 41.2 m
The radius of the sphere is r=5.15 cm=0.0515 m, and its surface is given by
So the total charge on the surface of the sphere is, using the charge density
:
The electrostatic force between the sphere and the point charge is:
where
ke is the Coulomb's constant
Q is the charge on the sphere
is the point charge
r is their separation
Re-arranging the equation, we can find the separation between the sphere and the point charge:
So the total charge on the surface of the sphere is, using the charge density
The electrostatic force between the sphere and the point charge is:
where
ke is the Coulomb's constant
Q is the charge on the sphere
r is their separation
Re-arranging the equation, we can find the separation between the sphere and the point charge: