A wire of length L, (=20, cm), is bent into a semicircular arc. If the two equal halves of the arc were each to be uniformly charged with charges ± Q,,, [ | Q | = 10 3 0 ] Coulomb where 0 is the permittivity (in SI, units) of free space] the net electric field at the centre O of the semicircular arc would be

(A) \left( {50 \times {{10}^3}\,N/C} \right)\hat j

(B) \left( {50 \times {{10}^3}\,N/C} \right)\hat i

(C) \left( {25 \times {{10}^3}\,N/C} \right)\hat j

JEE Mains · Physics · STD 12 - Electric charges and fields

A wire of length \(L\, (=20\, cm)\), is bent into a semicircular arc. If the two equal halves of the arc were each to be uniformly charged with charges \( \pm Q\,,\,\left[ {\left| Q \right| = {{10}^3}{\varepsilon _0}} \right]\) Coulomb where \(\varepsilon _0\) is the permittivity (in \(SI\, units\)) of free space] the net electric field at the centre \(O\) of the semicircular arc would be

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Verified Solution

Correct Answer

(D) \(\left( {25 \times {{10}^3}\,N/C} \right)\hat i\)

Detailed explanation

Given: Length of wire \(L=20 \,\mathrm{cm}\) charge \(Q=10^{3} \varepsilon_{0}\)

We know, electric field at the centre of the semicircular arc

\({E=\frac{2 K \lambda}{r}}\)

\({\text{ or, }}\quad E = \frac{{2K\left( {\frac{{2Q}}{{\pi r}}} \right)}}{r}\left[ {As\,\lambda = \frac{{2Q}}{{\pi r}}} \right]\)

\(=\frac{4 K Q}{\pi r^{2}}=\frac{4 K Q \pi^{2}}{\pi L^{2}}=\frac{4 \pi K Q}{L^{2}}=25 \times 10^{3} \,N / C \hat{i}\)

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