SATHEE: Electrostatics Question 245

Electrostatics Question 245

Question: Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in [IIT-JEE (Screening) 2001]

Options:

A) $\frac{q}{2 π^{2} ε_{0} R^{2}}$

B) $\frac{q}{4 π^{2} ε_{0} R^{2}}$

C) $\frac{q}{4 π ε_{0} R^{2}}$

D) $\frac{q}{2 π ε_{0} R^{2}}$

Show Answer

Answer:

Correct Answer: B

Solution:

$d l = R d θ$ ; Charge on $d l = λ R d θ$

$λ = \frac{q}{π R}$

Electric field at centre due to dl is $d E = k . \frac{λ R d θ}{R^{2}}$ .

We need to consider only the component $d E \cos θ,$ as the component dE sinq will cancel out because of the field at C due to the symmetrical element dl¢.

Total field at centre $= 2 \int_{0}^{π / 2} d E \cos θ$

$= \frac{2 k λ}{R} \int_{0}^{π / 2} \cos θ d θ = \frac{2 k λ}{R} = \frac{q}{2 π^{2} ε_{0} R^{2}}$

Alternate method : As we know that electric field due to a finite length charged wire on it’s perpendicular bisector is given by $E = \frac{2 k λ}{R} \sin θ .$

If it is bent in the form of a semicircle then $θ = 90^{\circ}$

therefore $E = \frac{2 k λ}{R}$ = $2 \times \frac{1}{4 π ε_{0}} (\frac{q / π R}{R})$ = $\frac{q}{2 π^{2} ε_{0} R^{2}}$

Electrostatics Question 245

Question: Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in [IIT-JEE (Screening) 2001]

Options:

Answer:

Solution:

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Electrostatics Question 245

Question: Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in [IIT-JEE (Screening) 2001]

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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