Answer:
option B
Explanation:
we know,
Force between two charge is calculated by
[tex]F = \dfrac{kq_1q_2}{r^2}[/tex]
r is the distance between the two charges
k is Coulomb constant.
give two small object with charge Q
now, Force
[tex]F = \dfrac{kQQ}{r^2}[/tex]......(1)
now,
New force if another charge is equal to 4 Q
[tex]F' = \dfrac{kQ(4Q)}{r^2}[/tex]
[tex]F' = 4\dfrac{kQQ}{r^2}[/tex]
now from equation 1
[tex]F' = 4 F[/tex]
Hence, the correct answer is option B
The magnitude of the force on Q when one of the charges Q is replaced with another charge 4Q is 4F (Option B)
Coulombs law of electricity states that the magnitude of the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance separating the two charges.
Mathematically, F = [tex]\frac{KQ1Q2}{d^2}[/tex]
If the initial charges on the two objects are each equal to Q and the distance between the charges is d, the magnitude of the force between the objects, F is given by the equation of coulombs law above i.e. F = [tex]\frac{KQQ}{d^2}[/tex]
If one of the charges is replaced with a charge 4Q, the force between the charges will now be given as:
F' = [tex]\frac{KQ4Q}{d^2}[/tex]
F' = 4 * [tex]\frac{KQQ}{d^2}[/tex]
Comparing the initial and final magnitude of the force on the objects:
F' = 4F
Therefore, the magnitude of the force on the Q now is 4F
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