The distance between charges 5×10−11C and −2.7×10−11C is 0.2m. The d

Question

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

Answer: Option C
:
C
If two opposite charges are separated by a certain distance, then for it's equalibrium 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^{- 11} C

then for it's
equalibrium

| F_{1} | = | F_{2} |

The Distance Between Charges 5×10−11C And −2.7×10−...

\Rightarrow \frac{K Q_{1} q}{(x + 0.2)^{2}} = \frac{K Q_{2} q}{x^{2}} \Rightarrow x = 0.556 m

(H e r e K = \frac{1}{4 π ϵ_{0}} a n d Q_{1} = 5 \times 10^{- 11} C, Q_{2} = - 2.7 \times 10^{- 11} C)

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