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The force acting between two point charges kept at a certain distance is 5 N. Now the magnitudes of the charges are doubled and the distance between them is halved. The force acting between them is ______ N.

Asked in GSEB Board July 2017 · Coulomb's law and superposition

Answer: (4) 80

Step-by-step solution

Given: F=5 N; each charge doubled; distance halved.

Idea: Coulomb's law, F=(kq₁q₂)/(r²). The force is proportional to the product of the charges and to 1/(r²).

Charges: q₁q₂→(2q₁)(2q₂)=4q₁q₂, a factor of 4.

Distance: r→r/2, so r²→(r²)/4 and 1/(r²) grows by another factor of 4.

F'=4×4× F=16×5=80 N

So the new force is 80 N.

Why the other options are wrong

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