Respuesta :
The force exerted between S and q is ___4 times____ the force exerted between S and p.
What is Coulomb's law ?
Coulomb's law states that , the electrostatic force of attraction or repulsion between two charged bodies is directly proportional to the product of the charges and is inversely proportional to the square of the distance between them
It is given by
[tex]\rm F = k_e \dfrac{q_1 q_2}{r^2}[/tex]
r is the distance between two bodies
It is given that
Charge q is 1 unit of distance away from the source charge S
Charge p is two times further away from the source charge S
force exerted between S and q
distance between the charges is 1 unit
[tex]\rm F_{qs} = k_e \dfrac{q_q q_s}{1^2}\\\\\rm F_{ps} = k_e \dfrac{q_p q_s}{2^2}\\[/tex]
The force exerted between S and q is _______ the force exerted between S and p.
[tex]\rm F_{qs} = k_e \dfrac{q_q q_s}{1^2} = F_{ps} = k_e \dfrac{q_p q_s}{2^2}\\\\[/tex]
Assuming the charges to be q units
Then
[tex]\rm F_{qs} = k_e \dfrac{q_q q_s}{1^2} = F_{ps} = k_e \dfrac{q_p q_s}{2^2}\\\\\\\\\dfrac {F_{qs}}{F_{ps}} = \dfrac{2^2}{1^2} = 4 \\\\\\{F_{qs}} = 4{F_{ps}}[/tex]
Therefore ,
The force exerted between S and q is ___4 times____ the force exerted between S and p.
To know more about Coulomb's Law
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