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The distance between charges 5 \times 10^{-11 }C and -2.7 \times 10^{-11 }C is 0.2 m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is 

 

  • Option 1)

    0.44 m 

  • Option 2)

    0.65 m

  • Option 3)

    0.556 m

  • Option 4)

    0.350 m

 

Answers (1)

As we have learnt, 

 

Neutral Point due to two unlike Point -

\dpi{100} \frac{Q_{1}}{Q_{2}}=\left ( \frac{x}{x+l} \right )^{2}

- wherein

Where, x = distance between Q1 and Q2

 

If two opposite charges are separated by a certain distance, then for it’s equilibrium a third charge should be kept outside and near the charge which is smaller in magnitude. 
Here, suppose third charge q is placed at a distance x from – 2.7 \times 10–11C then for it’s equilibrium |F1| = |F2|

\\*\Rightarrow \frac{kQ_1q}{(x+ 0.2)^2} = \frac{kQ_2q}{x^2} \Rightarrow x = 0.556 m \\*\left(Here\; k =\frac{1}{4\pi\epsilon_0}\;and\;Q_1 = 5\times 10^{-11}, Q_2 = 2.7\times 10^{-11}\right )


Option 1)

0.44 m 

Option 2)

0.65 m

Option 3)

0.556 m

Option 4)

0.350 m

Posted by

Vakul

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