Question 689: Electrostats & Capacitance

Question

Three charges $$ - {q_1}, + {q_2}$$ and $$ - {q_3}$$ are placed as shown in the figure. The $$x$$-component of the force on $$ - {q_1}$$ is proportional to

Accepted Answer

Force on charge $${q_1}$$ due to $${q_2}$$ is $${F_{12}} = k{{{q_1}{q_2}} \over {{b^2}}}$$
Force on charge $${q_1}$$ due to $${q_3}$$ is $${F_{13}} = k{{{q_1}{q_3}} \over {{a^2}}}$$
The $$X$$-component of the  force $$\left( {{F_x}} ight)$$ on
$$q{}_1$$ is $${F_{12}} + {F_{13}}$$ $$\sin 	heta $$
$$	herefore$$ $${F_x} = k{{{q_1}{q_2}} \over {{b^2}}} + k{{{q_1}{q_2}} \over {{a^2}}}\sin 	heta $$
$$	herefore$$ $${F_x} \propto {{{q_2}} \over {{b^2}}} + {{{q_3}} \over {{a^2}}}\sin 	heta $$

Answer

$${{{q_2}} \over {{b^2}}} - {{{q_3}} \over {{a^2}}}\cos 	heta $$

Answer

$${{{q_2}} \over {{b^2}}} + {{{q_3}} \over {{a^2}}}\sin 	heta $$

Answer

$${{{q_2}} \over {{b^2}}} + {{{q_3}} \over {{a^2}}}\cos 	heta $$

Answer

$${{{q_2}} \over {{b^2}}} - {{{q_3}} \over {{a^2}}}sin	heta $$

Electrostats & Capacitance

Three charges $ - {q_1}, + {q_2} and - {q_3} are placed as shown in the figure. The x-component of the force on - {q_1}$ is proportional to

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