A wire of length L=20 cm is… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Electromagnetism

2015mcqhard

A wire of length $L=20 cm$ is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges $+Q$ and $-Q$ as shown in the figure. The magnitude of the charge on each half is $|Q|={10}^{3}{\epsilon }_{0}$ , where ${\epsilon }_{0}$ is the permittivity of free the space. The net electric field at the centre $O$ is

JEE Main 2015 Physics, Electrostatics — question figure

Official previous-year question

Held on 11 Apr 2015 · Verified 6 Jul 2026.

Options

A
$(25\times {10}^{3})\hat{i}N{C}^{-1}$
B
$(50\times {10}^{3})\hat{i}N{C}^{-1}$
C
$(25\times {10}^{3})\hat{j}N{C}^{-1}$
D
$(50\times {10}^{3})\hat{j}N{C}^{-1}$

Did you get this right?

Solution

JEE Main 2015 Physics, Electrostatics — solution figure 1

$L=\pi R$

$\Rightarrow R=\frac{L}{\pi } =\frac{ 20}{ 100\pi } \text{m}=\frac{1}{ 5\pi } \text{m}$

$\Rightarrow$ due to a charge arc, electric field at centre is given by

$E=\frac{ 2K\lambda }{R} \mathrm{sin}\frac{\theta }{2}$

JEE Main 2015 Physics, Electrostatics — solution figure 2

${E}_{1}={E}_{2}=\frac{2k⋋}{R}\mathrm{sin}⁡\frac{90}{2} {⋋=\frac{Q}{\pi R/2}}$

undefined

Component along $\hat{j}$ gets cancelled and

${E}_{ net} =\sqrt{ 2 } {E}_{1}$

$=\frac{ 4KQ}{ \pi {R}^{2} }$

$=\frac{4\times 1({10}^{3}{\epsilon }_{0})}{4\pi {\epsilon }_{0} \pi {R}^{2}}=\frac{4\times {10}^{3}{\epsilon }_{0}}{4{\pi }^{2}{\epsilon }_{0}{R}^{2}}=\frac{{10}^{2}}{{R}^{2}}=\frac{100}{{(\frac{1}{5\pi })}^{2}}=25\times {10}^{3}$

${\vec{E}}_{net}=25\times {10}^{3}N{C}^{-1} \hat{i}$

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Solution

JEE Main 2015 Physics, Electrostatics — solution figure 1

$L=\pi R$

$\Rightarrow R=\frac{L}{\pi } =\frac{ 20}{ 100\pi } \text{m}=\frac{1}{ 5\pi } \text{m}$

$\Rightarrow$ due to a charge arc, electric field at centre is given by

$E=\frac{ 2K\lambda }{R} \mathrm{sin}\frac{\theta }{2}$

JEE Main 2015 Physics, Electrostatics — solution figure 2

${E}_{1}={E}_{2}=\frac{2k⋋}{R}\mathrm{sin}⁡\frac{90}{2} {⋋=\frac{Q}{\pi R/2}}$

undefined

Component along $\hat{j}$ gets cancelled and

${E}_{ net} =\sqrt{ 2 } {E}_{1}$

$=\frac{ 4KQ}{ \pi {R}^{2} }$

${\vec{E}}_{net}=25\times {10}^{3}N{C}^{-1} \hat{i}$

Options

Did you get this right?

Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism