Electrostatics Question 451
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 [Kerala PET 2002]
Options:
A) 0.44 m
B) 0.65 m
C) 0.556 m
D) 0.350 m
Show Answer
Answer:
Correct Answer: C
Solution:
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^{11}C $ then for it’s equilibrium $ | F _{1} |=| F _{2} | $
therefore $ \frac{kQ _{1}q}{{{(x+0.2)}^{2}}}=\frac{kQ _{2}q}{x^{2}} $
therefore x = 0.556 m $ ( \text{Here }k=\frac{1}{4\pi {\varepsilon _{0}}}\text{and}Q _{1}=5\times {{10}^{-11}}C,Q _{2}=-2.7\times {{10}^{-11}}C ) $
