SATHEE: Electrostatics Question 262

Electrostatics Question 262

Question: Four charges equal to -q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is [AIEEE 2004]

Options:

A) Only $x = \sqrt{2} a$

B) Only $x = - \sqrt{2} a$

C) Both $x = \pm \sqrt{2} a$

D) $x = \frac{3 a}{2}$ only

Show Answer

Answer:

Correct Answer: B

Solution:

Suppose electric field is zero at a point P lies at a distance d from the charge + Q.

At P $\frac{k Q}{d^{2}} = \frac{k (2 Q)}{{(a + d)}^{2}}$

therefore $\frac{1}{d^{2}} = \frac{2}{{(a + d)}^{2}}$

therefore $d = \frac{a}{(\sqrt{2} - 1)}$

Since d > a i.e. point P must lies on negative x-axis as shown at a distance x from origin hence $x = d - a$

$= \frac{a}{(\sqrt{2} - 1)} - a = \sqrt{2} a .$

Actually P lies on negative x-axis so $x = - \sqrt{2} a$

Electrostatics Question 262

Question: Four charges equal to -q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is [AIEEE 2004]

Options:

Answer:

Solution:

Engineering Preparation

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Syllabus

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Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Electrostatics Question 262

Question: Four charges equal to -q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is [AIEEE 2004]

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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