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  3. Can someone explain - Electrostatics - JEE Main-2
# Engineering

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

  • Option 1)

    \frac{q_{2}}{b^{2}}-\frac{q_{3}}{a^{2}}\cos \Theta

  • Option 2)

    \frac{q_{2}}{b^{2}}+\frac{q_{3}}{a^{2}}\sin \Theta

  • Option 3)

    \frac{q_{2}}{b^{2}}+\frac{q_{3}}{a^{2}}\cos \Theta

  • Option 4)

    \frac{q_{2}}{b^{2}}-\frac{q_{3}}{a^{2}}\sin \Theta

 

Answers (2)
A Aadil

As we learnt in

Magnitude of the Resultant force -

F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }

- wherein

 

 

F\! orce \: \: on\: \: \left ( -q_{1} \right )due\: \:\: to\: q_{2}= \frac{-q_{1}q_{2}}{4\pi \varepsilon _{0}b^{2}}

\therefore \: \: \: F_{1}= \frac{q_{1}q_{2}}{4\pi \varepsilon _{0}b^{2}}\: along\: (q_{1}q_{2})

Force\: \: on\: (-q_{1})\: \: due\: \: to\: \: \left ( -q_{3} \right )= \frac{\left (-q_{1} \right )\left ( -q_{3} \right )}{4\pi\varepsilon _{0}a^{2} }

F_{2}= \frac{q_{1}q_{3}}{4\pi\varepsilon _{0}a^{2}}\: as\: \: shown\:

F2   makes an angle of \left ( 90^{\circ}-\Theta \right )\: with\ direction\ of\ line\ joining \left ( q_{1}q_{2} \right )

Resolved part of F2   along q1q2

= F_{2}\cos \left ( 90^{\circ}-\Theta \right )

= \frac{q_{1}q_{3}\sin \Theta }{4\pi \varepsilon _{0}a^{2}}\: \: along\: \: \left ( q_{1}q_{2} \right )

\therefore \: \: total\: force \: on\: \left ( -q_{1} \right ) = \left [ \frac{q_{1}q_{2} }{4\pi \varepsilon _{0}b^{2}}+ \frac{q_{1}q_{3}\sin \Theta }{4\pi \varepsilon _{0}a^{2}} \right ]\: \: along\: x\: \: axis

\therefore \: \: \: component \: \: of \: \: force\: \: \alpha \left [ \frac{q_{2}}{b^{2}} +\frac{q_{3}}{a^{2}}\sin \Theta \right ]

 


Option 1)

\frac{q_{2}}{b^{2}}-\frac{q_{3}}{a^{2}}\cos \Theta

Incorrect

Option 2)

\frac{q_{2}}{b^{2}}+\frac{q_{3}}{a^{2}}\sin \Theta

Correct

Option 3)

\frac{q_{2}}{b^{2}}+\frac{q_{3}}{a^{2}}\cos \Theta

Incorrect

Option 4)

\frac{q_{2}}{b^{2}}-\frac{q_{3}}{a^{2}}\sin \Theta

Incorrect

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