Three identical charges,each 2 mu C,lie at th | vedclass

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(C) From the figure,the net force $F_{	ext{net}}$ due to $F_1$ and $F_2$ makes an angle $	heta$ with force $F_2$.Since $F_1$ and $F_2$ are perpendic

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$	an ^{-1}\left(\frac{16}{9}ight)$

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(C) From the figure,the net force $F_{	ext{net}}$ due to $F_1$ and $F_2$ makes an angle $	heta$ with force $F_2$.Since $F_1$ and $F_2$ are perpendicular to each other,the angle $	heta$ is given by $	an 	heta = \frac{F_1}{F_2}$.Using Coulomb's Law,the force $F_1$ on charge at $B$ due to charge at $A$ is $F_1 = k \cdot \frac{q_A q_B}{(AB)^2} = k \cdot \frac{q^2}{(3)^2}$.The force $F_2$ on charge at $B$ due to charge at $C$ is $F_2 = k \cdot \frac{q_C q_B}{(BC)^2} = k \cdot \frac{q^2}{(4)^2}$.Therefore,$	an 	heta = \frac{F_1}{F_2} = \frac{k \cdot q^2 / 9}{k \cdot q^2 / 16} = \frac{16}{9}$.Thus,$	heta = 	an ^{-1}\left(\frac{16}{9}ight)$.

Three identical charges,each $2 \mu C$,lie at the vertices of a right-angled triangle as shown in the figure. Forces on the charge at $B$ due to the charges at $A$ and $C$ respectively are $F_1$ and $F_2$. The angle between their resultant force and $F_2$ is

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