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lyudmila [28]
3 years ago
13

Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original v

alue and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them
Physics
1 answer:
Gelneren [198K]3 years ago
6 0

Answer:

F' = F/12

Therefore, the electrostatic force is reduced to one-twelve of its original value.

Explanation:

The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:

F = kq₁q₂/r²   ---------- equation 1

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between charges

Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:

F' = kq₁'q₂'/r'²

where,

q₁' = (1/3)q₁

q₂' = q₂

r' = 2r

Therefore,

F' = k(1/3 q₁)(q₂)/(2r)²

F' = (1/12)kq₁q₂/r²

using equation 1:

<u>F' = F/12</u>

<u>Therefore, the electrostatic force is reduced to one-twelve of its original value.</u>

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