Question

Chapter 18, Problem 14 GO The dranings show three charges that have the same magnitude but may haive different signs. In all cases the distance d between the charges is the same. The magnitude of the charges is lo-7.8 C, and the distance between them is ฮ«J 8 mm. Determine the magnitude of the net force on charge 2 foreach of the tr ee e ing. (o) F- (b) F- (c) F-

Answer 1

Given $q=7.8\mu C=7.8*10^{-6}C$

$d=3.8mm=0.0038m$

(a)q₂ is positive .q₁ is negative,so pulling q₂ towards left and q₃ is pushing q₂ towards left

Add the Forces

$F_{1}=kq_{1}q^{2}/d^{2}$

$F_{1}=9*10^{9}*7.8*10^{-6}C*7.8*10^{-6}C/(0.0038m)^{2}$

$F_{1}=9*10^{9}*7.8*10^{-6}*7.8*10^{-6}/0.0038^{2}$

$F_{1}=0.54756/0.0038^{2}$

$F_{1}=37919.67N$

since charges are identical and distance is same

$F_{3}=F_{1}=37919.67N$

Net Force $F=F_{1}+F_{3}=2F_{1}=2*37919.67N$

ANSWER: ${\color{Red} F=75839.34N}$

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b)

all charges are positive, so forces are repulsive

q₁ is pushing q₂ towards right and q₃ is pushing q₂ towards left

since the forces are equal and opposite,they cancel out

Net force is zero

ANSWER: ${\color{Red} F=0N}$

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c)

q₂ is negative .q₁ is positive ,so pulling q₂ towards left

q₃ is positive ,so pulling q₂ downwards

From (a)$F_{1}=37919.67N$ (towards left)

$F_{3}=37919.67N$ (downwards)

Net Force ,$F=\sqrt{F_{1}^{2}+F_{3}^{2}}$

$F=\sqrt{37919.67^{2}+37919.67^{2}}$

ANSWER: ${\color{Red} F=53626.51N}$

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