JEE Mains · Physics · STD 12 - 1. 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
- A \(\left( {50 \times {{10}^3}\,N/C} \right)\hat j\)
- B \(\left( {50 \times {{10}^3}\,N/C} \right)\hat i\)
- C \(\left( {25 \times {{10}^3}\,N/C} \right)\hat j\)
- D \(\left( {25 \times {{10}^3}\,N/C} \right)\hat i\)
Answer & Solution
Correct Answer
(D) \(\left( {25 \times {{10}^3}\,N/C} \right)\hat i\)
Step-by-step Solution
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 \({E=\frac{2 K \lambda}{r}}\)…
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