JEE Main 2015 · 11 Apr · Shift 1 · Physics Q14

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PhysicsSingle correct+4 / -1Question 14 of 90

A wire, of length L(=20 cm), is bent into a semi-circular arc. If the two equal halves, of the arc, were each to be uniformly charged with charges ±Q, [|Q| = $10^3\epsilon_0$ Coulomb where $\epsilon_0$ is the permittivity (in SI units) of free space] the net electric field at the centre O of the semi-circular arc would be :

Axes X and Y. A semicircular arc above its centre O: the left quarter carries + charges and the right quarter − charges.

(50\times10^3\text{ N/C})\,\hat j

(25\times10^3\text{ N/C})\,\hat i

(25\times10^3\text{ N/C})\,\hat j

(50\times10^3\text{ N/C})\,\hat i

Answer and worked solution

Official answer: B

$R=\dfrac L\pi=\dfrac{0.2}\pi$ m. Each quarter arc carries charge Q with $\lambda=\dfrac{2Q}{\pi R}$ , and gives $\dfrac{\sqrt2k\lambda}R$ along its bisector.

The + quarter (upper left) pushes the field towards the lower right. The − quarter (upper right) pulls it towards the upper right. The vertical parts cancel and the horizontal parts add:
$E=\frac{2k\lambda}R=\frac{Q}{\pi^2\epsilon_0R^2}=\frac{10^3}{0.04}=25\times10^3\text{ N/C along }\hat i.$

Topics:Electric field Coulomb's law and superposition

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 14