A point charge Q= 4.60 uC is held fixed at the origin. A second point charge q=1.20 uC with mass of 2.80 * 10^-4 is placed on the x-axis 0.250 m away from the origin.
A point charge 𝑄 = +51.1 µ𝐶 is held fixed at the origin. A second point charge 𝑄 = −1.91 µ𝐶 and their mass 𝑚 = 0.0304 𝑘𝑔 is placed at the location (x=0.432 m, y=0 m)
a) Find the electric potential energy of this system of charges
b) If the second charge is released from rest, find the speed of the second charge when it reaches the point (x=0.182 m, y=0 m).
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To find the electric force experienced by the charged particle due to the fixed point charge, we can use Coulomb's Law.
Coulomb's Law states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * (|Q| * |q|) / r^2
where F is the electric force, k is the Coulomb's constant (k = 8.99 * 10^9 N*m^2/C^2), |Q| and |q| are the magnitudes of the two charges, and r is the distance between the charges.
In this case, Q = 4.60 μC, q = 1.20 μC, and r = 0.250 m.
Converting the charges from microcoulombs to coulombs, we have:
Q = 4.60 μC * 10^(-6) C/μC = 4.60 * 10^(-6) C
q = 1.20 μC * 10^(-6) C/μC = 1.20 * 10^(-6) C
Plugging these values into the formula, we get:
F = (8.99 * 10^9 N*m^2/C^2) * (4.60 * 10^(-6) C) * (1.20 * 10^(-6) C) / (0.250 m)^2
Calculating this expression will give us the electric force experienced by the charged particle.