SATHEE: Electrostatics 1 Question 10

Electrostatics 1 Question 10

10. 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

(1987, 2M)

(a) $- \frac{Q}{2}$

(b) $- \frac{Q}{4}$

(d) $+ \frac{Q}{2}$

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Answer:

Correct Answer: 10. (b)

Solution:

Since, $q$ is at the centre of two charges $Q$ and $Q$ , net force on it is zero, whatever the magnitude and sign of charge

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on it. For the equilibrium of $Q, q$ should be negative because other charge $Q$ will repel it, so $q$ should attract it. Simultaneously these attractions and repulsions should be equal.

$\frac{1}{4 π ε_{0}} \frac{Q Q}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{Q q}{(r / 2)^{2}}$

or with sign

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

Electrostatics 1 Question 10

10. 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

Answer:

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Electrostatics 1 Question 10

10. 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

Answer:

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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10. 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