SATHEE: Electrostatics Question 321

Electrostatics Question 321

Question: A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium, if $q$ is equal to [IIT 1987; CBSE PMT 1995; Bihar MEE 1995; CPMT 1999; MP PET 1999; MP PMT 1999, 2000; RPET 1999; KCET 2001; AIEEE 2002; AFMC 2002; Kerala PMT 2004; J & K CET 2004]

Options:

A) $- \frac{Q}{2}$

B) $- \frac{Q}{4}$

C) $+ \frac{Q}{4}$

D) $+ \frac{Q}{2}$

Show Answer

Answer:

Correct Answer: B

Solution:

Equilibrium of charge B is considered.

Hence for it’s equilibrium $| F_{A} | = | F_{C} |$

therefore $\frac{1}{4 π ε_{0}} \frac{Q^{2}}{4 x^{2}} =$

$\frac{1}{4 π ε_{0}} \frac{q Q}{x^{2}}$

therefore $q = \frac{- Q}{4}$

Short Trick: For such type of problem the magnitude of middle charge can be determined if either of the extreme charge is in equilibrium by using the following formula.

If charge A is in equilibrium then q = ? $Q_{B} {(\frac{x_{1}}{x})}^{2}$

If charge B is in equilibrium then $q = - Q_{A} {(\frac{x_{2}}{x})}^{2}$

If the whole system is in equilibrium then use either of the above formula.

Electrostatics Question 321

Options:

Answer:

Solution:

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Electrostatics Question 321

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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