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{{\pi }^{2}}{\varepsilon _{0}}R^{2}} $
B) $ \frac{q}{4{{\pi }^{2}}{\varepsilon _{0}}R^{2}} $
C) $ \frac{q}{4\pi {\varepsilon _{0}}R^{2}} $
D) $ \frac{q}{2\pi {\varepsilon _{0}}R^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ dl=Rd\theta $ ; Charge on $ dl=\lambda Rd\theta $
$ { \lambda =\frac{q}{\pi R} } $
Electric field at centre due to dl is $ dE=k.\frac{\lambda Rd\theta }{R^{2}} $ .
We need to consider only the component $ dE\cos \theta , $ 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}^{\pi /2}{dE\cos \theta } $
$ =\frac{2k\lambda }{R}\int _{0}^{\pi /2}{\cos \theta d\theta }=\frac{2k\lambda }{R}=\frac{q}{2{{\pi }^{2}}{\varepsilon _{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{2k\lambda }{R}\sin \theta . $
If it is bent in the form of a semicircle then $ \theta =\text{ 9}0{}^\circ $
therefore $ E=\frac{2k\lambda }{R} $ = $ 2\times \frac{1}{4\pi {\varepsilon _{0}}}( \frac{q/\pi R}{R} ) $ = $ \frac{q}{2{{\pi }^{2}}{\varepsilon _{0}}R^{2}} $
