Answer to Question #250068 in Physics for Nene

Question #250068

Two-point charges are attracted from each other with a force of 6kN. If the distance between them is reduced to one fourth of its original value, determine the new force of attraction between them.


1
Expert's answer
2021-10-12T12:33:00-0400

According to the Coulomb's law, the force between two point charges is proportional to the squared distance between them:


F11r12F_1\sim\dfrac{1}{r_1^2}

where F1=6kNF_1 = 6kN is the force and r1r_1 is the distance in first case. Similarly:


F21(r1/4)2F_2\sim\dfrac{1}{(r_1/4)^2}

where F2F_2 is the new force and r1/4r_1/4 is the reduced distance. Taking the ration of these two expressions, obtain:


F2F1=r12(r1/4)2F2=F1r12(r1/4)2=F1r1242r12=16F1F2=6kN16=96kN\dfrac{F_2}{F_1} = \dfrac{r_1^2}{(r_1/4)^2}\\ F_2 = F_1\dfrac{r_1^2}{(r_1/4)^2} = \dfrac{F_1r_1^2\cdot 4^2}{r_1^2} = 16F_1\\ F_2 = 6kN\cdot 16 = 96kN

Answer. 96kN.


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