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Two positive point charges, each with charge q, separated by a distance d, repel each other with a force of magnitude 20 N. What is the magnitude of the force between two positive point charges of magnitude 11.11 q, separated by a distance 2.5 d in units of N

1 Answer

2 votes

Answer:

The value is
F_2 = 395 \ N

Step-by-step explanation:

From the question we are told that

The magnitude of the charge of each positive charge for the first case is
q_1 = q_2 = q

The distance between the charges for the first case is
d

The force between the charges for the first case is
F = 20 \ N

The magnitude of the charge of each positive charge for the second case is
q_1 = q_2=11.11q

The distance between the charge for the second case is
2.5d

Generally for the first case the force between the charge is mathematically represented as


F_1 = (k * q^2 )/(d^2)

Where k is the Coulomb constant with value
k = 9*10^(9) \ kg\cdot m^3\cdot s^(-4) \cdot A^(-2).

So


20= (k * q^2 )/(d^2)

Generally for the second case the force between the charge is mathematically represented as


F_2 = (k * (11.11q)^2)/((2.5d)^2)


F_2 = (k * 11.11^2 *q^2)/(2.5^2d^2)

=>
F_2 = (11.11^2)/(2.5^2) F_1

=>
F_2 = (11.11^2)/(2.5^2) * 20

=>
F_2 = 395 \ N

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