Question #280228
  1. a Calculate the force between two charges q1 of

2.0 μC and q2 4.0 μC separated by r = 5.0 cm. (Ans: 29 N)


b Let the force calculated in a be F. In terms of F

and without further calculations, state the force

between these charges when:


i the separation r of the charges is doubled (Ans: F/4)

ii q1 and r are both doubled (F/2)

iii q1, q2 and r are all doubled. (F)


Please explain how these problems were solved.


1
Expert's answer
2021-12-16T17:58:21-0500

(a)

Force

F=kq1q2r2F=\frac{kq_1q_2}{r^2}


F=9×109×2×106×4×1060.052=28.8NF=28.8N(approximatelyF=29N)F=\frac{9\times10^{9}\times2\times10^{-6}\times4\times10^{-6}}{0.05^2}=28.8N \\F=28.8N \\ (approximately F=29N)

(1)

Force

F=kq1q2r12F=\frac{kq_1q_2}{r_1^2}

r1=2rr_1=2r

F=kq1q2(2r)2=F4F=\frac{kq_1q_2}{(2r)^2}=\frac{F}{4}


(2)

F=kq1q2r12F=\frac{kq_1q_2}{r_1^2}

q1=2q1,r=2rq_1=2q_1,r=2r

F=2kq1q2(2r)2=kq1q22r2=F2F=\frac{2kq_1q_2}{(2r)^2}=\frac{kq_1q_2}{2r^2}=\frac{F}{2}

(3)

F=kq1q2r12F=\frac{kq_1q_2}{r_1^2}

q1=2q1,q2=2q2,r=2rq_1=2q_1,q_2=2q_2,r=2r

F=k2q12q2(2r)2F=4kq1q24r2=FF=\frac{k2q_12q_2}{(2r)^2} \\F=\frac{4kq_1q_2}{4r^2}=F

F=kq1q2r2F=\frac{kq_1q_2}{r^2}


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