Question

Three electric charges are located as shown in the diagram, with q-7.8.uC, q7.8.uC, q8.7.pC, a b = 0.62-m, and c 0.77-m. What is the electric force on charge q due to charges qi and q? 0.38 m, 2. magnitude a. 0.805.N b. 0.7425.N C. 1.207 N d. 0.9494 N e 0.7474.N f. 1.061 N 3. direction (as rotation angle) a. 10.73 b. 12.13 c. 16.08 d. 23.76 e. 8.538 13.97 q1 a 42
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Answer #1

b 0.62m, Fa Fb a0.38.m, 0.86m 0.99m 51.2 b q2 63.7 qi a

Like charges attracts and unlike charges repels ,so identify the direction of forces

First Find Forces Fa and Fb

q = -7.8*10-6 C

q1 = -7.8*10-6 C

q2 = 8.7*10-6 C

By Coulomb's Law

F =kq_{1}q_{2}/r^{2} where k is 8.99*109

----------------------------

F_{a} =kq_{1}q/r^{2}

F_{a} =8.99*10^{9}*7.8*10^{-6}*7.8*10^{-6}/0.86^{2}

F_{a} =0.7395N

--------------------------

F_{b} =kq_{2}q/r^{2}

F_{a} =8.99*10^{9}*7.8*10^{-6}*8.7*10^{-6}/0.99^{2}

F_{a} =0.622N

-------------------------

Now split these forces into its Sine and Cosine components

Consider all Forces in x direction

F_{x}=F_{a}Cos63.7 +F_{b}Cos51.2

F_{x}=0.7395*Cos63.7 +0.622*Cos51.2

F_{x}=0.7174 N

-----------------------

Consider all Forces in y direction

F_{y}=F_{a}Sin63.7 -F_{b}Sin51.2

F_{y}=0.7395*Sin63.7 -0.622*Sin51.2

F_{y}=0.1782N

--------------------

Resultant Force F=\sqrt{F_{x}^{2}+F_{y}^{2}}

F=\sqrt{0.7174^{2}+0.1782^{2}}

F=0.7392N

ANSWER: Option b. 0.7425N

=======================

Direction

tano -F/FI

tan\theta = 0.1782/0.7174

\theta =13.948^{\circ}

ANSWER: Option f. 13.97 deg

=========================

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