Three point charges are placed along a straight line with successive charges being 3.0 m apart. The central one has a charge of −3.0 μC, whilst the outer two have charges of +3.0 μC each. If the charges are separated by a vacuum, calculate the magnitude of the total force on one of the outermost charges due to the other two.
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How I am proceeding:
Using 9*10^9 (|3*10^-6|*|3*10^-6|/3^2 + |3*10^-6|*|3*10^-6|/6^2)
But my answer is way out.
why do you have absolute signs on the charges. The sign of the charge is important.
force=kq(-q)/3^2 + kqq/6^2= kqq(-1/9+1/36)=kqq(-4/36+1/36)=kqq*3/36
where q=3e-6
To calculate the magnitude of the total force on one of the outermost charges, we can use Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
In this case, we have three charges arranged in a straight line, with the central charge being -3.0 μC and the outer charges being +3.0 μC each. The distance between the charges is given as 3.0 m.
To calculate the magnitude of the force on one of the outermost charges, let's call it Q1, due to the other two charges, Q2 and Q3, we can use the following formula:
F = k * |Q1| * |Q2| / r^2
where:
F is the magnitude of the force
k is the electrostatic constant, approximately 9 x 10^9 Nm^2/C^2
|Q1| and |Q2| are the magnitudes of the charges
r is the distance between the charges
In this case, Q1 = 3.0 μC, Q2 = Q3 = 3.0 μC, and r = 3.0 m.
Plugging in the values into the formula, we get:
F = (9 x 10^9 Nm^2/C^2) * (3.0 μC) * (3.0 μC) / (3.0 m)^2
Simplifying the calculation, we have:
F = (9 x 10^9) * (9 x 10^-12 C^2) / (3.0^2) m^2
F = 9 x 9 / 9 m^2
F = 9 N
Therefore, the magnitude of the total force on one of the outermost charges due to the other two is 9 Newtons.