The ratio of two charges is 2:3 and their distance of separation is 5cm. If the force of attraction between them is 96 dyn. Determine the magnitude of the charges? With calculation pls?

Question

Answer

Given:
Ratio of two charges = 2:3
Distance of separation = 5 cm
Force of attraction = 96 dyn

Solution:
Let the charges be Q1 and Q2. Therefore, their ratio is 2:3.
Let's assume that Q1 is 2x and Q2 is 3x.

Step 1:
Use Coulomb's Law to find the electric force between two charges.
F = (k * Q1 * Q2) / d^2
where k is Coulomb's constant (9 x 10^9 Nm^2/C^2)
d is the distance of separation between the charges

Step 2:
Substitute the given values in the equation
96 = (9 x 10^9 * 2x * 3x) / 0.05^2

Step 3:
Simplify the equation
96 = (54 x^2 * 10^9) / 0.0025
x^2 = (96 x 0.0025) / 54 x 10^9
x^2 = 0.000004444
x = 0.002107

Step 4:
Find the values of Q1 and Q2
Q1 = 2x = 0.004215 C
Q2 = 3x = 0.006322 C

Step 5:
Verify the answer by substituting the values of Q1 and Q2 in Coulomb's Law equation
F = (9 x 10^9 * 0.004215 * 0.006322) / 0.05^2
F = 96 dyn (As given in the question)

Therefore, the magnitude of the charges are:
Q1 = 0.004215 C
Q2 = 0.006322 C

Answer

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