Get Answers to all your Questions

header-bg qa

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q/q equals

  • Option 1)


  • Option 2)


  • Option 3)


  • Option 4)



Answers (2)


As we learnt in 

Principle of Super Position -

It states that total force acting on a given charge due to number of charges is the Vector sum of the individual force acting on that charge due to all the charges.

- wherein


 F=\tfrac{1}{4\pi\:\varepsilon _{o}}\frac{q^{2}}{(a^{2}+a^{2})}=\frac{q^{2}}{4\pi\:\varepsilon o(2a^{2})}

\therefore  total force of attraction along diagonal (taking \cos \Theta component)

=\frac{1}{4\pi\:\varepsilon \varepsilon _{o}}\:\left \{ \frac{Qq}{a^{2}\:} \frac{1}{\sqrt 2}+ \frac{Qq}{a^{2}\sqrt 2} \right \}= \frac{1}{4\pi \varepsilon _{o}}\left \{\frac{Qq\sqrt2}{a^{2}} \right \}

\frac{Q^{2}}{2a_{2}}=\frac{Qq\sqrt 2}{a^{2}}\\ \Rightarrow \frac{Q}{q}=-2\sqrt 2

Option 1)


This option is correct.

Option 2)


This option is incorrect.

Option 3)


This option is incorrect.

Option 4)


This option is incorrect.

Posted by


View full answer

Crack JEE Main with "AI Coach"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support