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  • Stuck here, help me understand: 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

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)

    -2\sqrt{2}

  • Option 2)

    -1

  • Option 3)

    1

  • Option 4)

    \frac{-1}{\sqrt{2}}

 

Answers (2)

best_answer

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)

-2\sqrt{2}

This option is correct.

Option 2)

-1

This option is incorrect.

Option 3)

1

This option is incorrect.

Option 4)

\frac{-1}{\sqrt{2}}

This option is incorrect.

Posted by

prateek

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