Electrostatics Question 720

Question: A thin glassrod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half, and a charge -Q is uniformly distributed along the lower half, as shown in fig. The electric field E at P, the center of the semicircle, is

Options:

A) Qπ2ε0r2

B) 2Qπ2ε0r2

C) 4Qπ2ε0r2

D) Q4π2ε0r2

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Take PO as the x-axis and PA as the y-axis, Consider two elements EF and Ef of width dθ at angular distance θ above and below PO, respectively.

The magnitude of the fields at P due to either element is dE=14πε0rdθ×Q/(πr/2)r2=Q2π2ε0r2dθ R

esolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PB. Therefore, field at P due to pair of elements is 2dEsinθ

E=0π/22dEsinθ

=20π/2Q2πε0r2sinθdθ=Qπ2ε0r2



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