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| = 103 ε 0 Coulomb where ε 0 is the permittivity (in SI units) of free space ] the net electric field at the center O of the semi-circular arc would be: (a) (50 × 103 N/C) j (b) (25 × 103 N/C) i (c) (25 × 103 N/C) j (d) (50 × 103 N/C) i

Answer: (b). +Q−QOE(+Q)E(−Q)E net (+x) π R = L ⇒ R = (0.2)/(π) m; each quarter arc has λ = Q/L/2 = 2Q/L. Quarter arc field: E₁ = (2kλsin 45°)/(R), along its bisector at 45° to the x-axis. For +Q and -Q halves the y-components cancel and x-components add: E = 2E₁cos 45° = (2kλ)/(R) = (Q)/(ε₀ L²). With Q = 10³ε₀: E = (10³)/((0.2)²) = 25×10³ N/C along i. So the answer is (b) (25×10³ N/C) i.

A wire, of length L(=20 cm), is bent into a semi-circular arc. If the… | JEE Main 2015 (11 Apr, Shift 1)

Q43JEE Main 2015 · 11 APR · SHIFT 1Difficult

Electric Field & Field Lines

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 $\pm$ Q, [ $|Q|$ = 103 $\varepsilon$ 0 Coulomb where $\varepsilon$ 0 is the permittivity (in SI units) of free space ] the net electric field at the center O of the semi-circular arc would be:

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Answer: (b)

$\pi R = L \Rightarrow R = \dfrac{0.2}{\pi}$ m; each quarter arc has $\lambda = \dfrac{Q}{L/2} = \dfrac{2Q}{L}$ .
Quarter arc field: $E_1 = \dfrac{2k\lambda\sin45^\circ}{R}$ , along its bisector at $45^\circ$ to the x-axis.
For $+Q$ and $-Q$ halves the y-components cancel and x-components add: $E = 2E_1\cos45^\circ = \dfrac{2k\lambda}{R} = \dfrac{Q}{\varepsilon_0 L^2}$ .
With $Q = 10^3\varepsilon_0$ : $E = \dfrac{10^3}{(0.2)^2} = 25\times10^3$ N/C along $\hat i$ .
So the answer is (b) $(25\times10^3\text{ N/C})\hat i$ .

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Electric Field & Field Lines question from JEE Main 2015 (11 Apr, Shift 1)

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