SATHEE: Electrostatics Question 451

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 $\frac{k Q_{1} q}{{(x + 0.2)}^{2}} = \frac{k Q_{2} q}{x^{2}}$

therefore x = 0.556 m $(Here k = \frac{1}{4 π ε_{0}} and Q_{1} = 5 \times 10^{- 11} C, Q_{2} = - 2.7 \times 10^{- 11} C)$

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Electrostatics Question 451

Question: The distance between charges 5×10−11C and −2.7×10−11C 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:

Answer:

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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]