three equal charges each of 2.0×10^-6 are fixed at three corner of an equilateral triangle of side 5cm .find the coulomb force experience due to other two charges
d = .05 meter
Fx = k [Q^2/.05^2]cos 30 from each
Fy cancels
so it is just 2Fx (repels of course because same sign)
I don't know the answer
To find the Coulomb force experienced by one of the charges due to the other two 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.
The formula for Coulomb's Law is:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges,
k is the electrostatic constant,
q1 and q2 are the charges, and
r is the distance between the charges.
In this case, since the charges are equal at each corner of an equilateral triangle, we can consider two charges (q1) and (q2), and calculate the force between them.
So, we have:
q1 = 2.0×10^-6 C (Each charge is 2.0×10^-6 C)
q2 = 2.0×10^-6 C
r = Side length of the equilateral triangle = 5 cm = 0.05 m (converted to meters)
Now, we can substitute the values into the formula and solve for the force (F).
F = k * (q1 * q2) / r^2
The value of k (the electrostatic constant) is 9 * 10^9 Nm^2/C^2.
Substituting the values:
F = (9 * 10^9 Nm^2/C^2) * [(2.0×10^-6 C) * (2.0×10^-6 C)] / (0.05 m)^2
Calculating the expression in the brackets first:
F = (9 * 10^9 Nm^2/C^2) * (4 * 10^-12 C^2) / (0.05 m)^2
Simplifying the equation:
F = (9 * 10^9 N * (4 * 10^-12) C) / (0.05^2 m^2)
F = (9 * 4 * 10^-3) * (10^9 N * C) / (0.0025 m^2)
F = 36 * 10^6 N * C / 0.0025 m^2
F = 36 * 10^6 * (C * N) / (m^2 / m^2)
F = 36 * 10^6 N
Therefore, the Coulomb force experienced by one of the charges due to the other two charges is 36 * 10^6 Newtons.