Two point charges q1 &q2 separated by a distance of 3.0 m experience a mutual force of 16x10^-15N.Calculate the magnitude of charge when q1=q2=q. What will be magnitude of force if seperation between the charges is changed to 6.0m?
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To calculate the magnitude of charge, we can use Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is represented as:
F = k * (q1 * q2) / r^2
Where:
F is the force between the charges,
k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2),
q1 and q2 are the magnitudes of the charges,
r is the separation distance between the charges.
Given:
q1 = q2 = q (charges are equal)
F = 16 x 10^-15 N
r = 3.0 m
Substituting these values into Coulomb's law, we can solve for q:
16 x 10^-15 = (9 x 10^9 * (q * q)) / (3.0^2)
Simplifying the equation:
16 x 10^-15 = (9 x 10^9 * q^2) / 9
Cross-multiplying:
16 x (9 / 10^15) = q^2
Dividing both sides by 144:
q^2 = (16 x (9 / 10^15)) / 144
Finally, taking the square root of both sides:
q = √((16 x (9 / 10^15)) / 144)
Calculating this expression will give us the magnitude of the charge.
Now, to answer the second part of your question: what will be the magnitude of force if the separation between the charges is changed to 6.0 m, we can use the same Coulomb's law equation by substituting the new separation distance r = 6.0 m into the formula. The charges q1 and q2 remain the same since their magnitudes were given as equal (q1 = q2 = q).
F = (9 x 10^9 * (q * q)) / (6.0^2)
Calculating this equation will give us the new magnitude of force.