Question

(13%) Problem 4: A charged particle (q-7.5 10-10 C) experiences a force of F-2.25i-6.2j N in an electric field 50% Part (a) Write an expression for the electric field vector E to which the charge is subject, in terms of the force F 50% Part b) Assume this field is generated by a point charge of Q-5x 109 C. How far away is this charge located? Give your answer in meters Grade Summa

Answer 1

Given $F=(2.25i-6.2j)N$

Use Formula $F=Eq$

$F/q=E$

$E=(2.25i-6.2j)/(7.5*10^{-10})$

$E=(0.3i-0.8267j)*10^{10}$

Part A) ANSWER: ${\color{Red} E=(3i-8.267j)*10^{9}N/C}$

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$E=kq/r^{2}$

$E=(3i-8.267j)*10^{9}N/C$

$E=8.7945*10^{9}N/C$

$8.7945*10^{9}N/C=k*5*10^{-9}C/r^{2}$

$8.7945*10^{9}=9*10^{9}*5*10^{-9}/r^{2}$

$8.7945*10^{9}=9*5/r^{2}$

$r^{2}=9*5/8.7945*10^{9}$

$r^{2}=45*10^{-9}/8.7945$

ANSWER: $r=7.15*10^{-5}m$

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