JEE MAIN - Physics (2021 - 27th August Evening Shift - No. 17)
Figure shows a rod AB, which is bent in a 120$$^\circ$$ circular arc of radius R. A charge ($$-$$Q) is uniformly distributed over rod AB. What is the electric field $$\overrightarrow E $$ at the centre of curvature O ?
_27th_August_Evening_Shift_en_17_1.png)
$${{3\sqrt 3 Q} \over {8\pi {\varepsilon _0}{R^2}}}(\widehat i)$$
$${{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}(\widehat i)$$
$${{3\sqrt 3 Q} \over {16{\pi ^2}{\varepsilon _0}{R^2}}}(\widehat i)$$
$${{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}( - \widehat i)$$
Explanation
$$\varepsilon = {{2k\lambda } \over R}\sin \left( {{\theta \over 2}} \right)( - \widehat i)$$
$$\lambda = \left( {{{ - Q} \over {R\theta }}} \right) = \left( {{{ - Q} \over {R.{{2\pi } \over 3}}}} \right)$$
$$\lambda = {{ - 3Q} \over {2\pi R}}$$
$$\varepsilon = {{2k} \over R}.{{ - 3Q} \over {2\pi R}}.\sin (60^\circ )( - \widehat i)$$
$$\varepsilon = {{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}( + \widehat i)$$
$$\lambda = \left( {{{ - Q} \over {R\theta }}} \right) = \left( {{{ - Q} \over {R.{{2\pi } \over 3}}}} \right)$$
$$\lambda = {{ - 3Q} \over {2\pi R}}$$
$$\varepsilon = {{2k} \over R}.{{ - 3Q} \over {2\pi R}}.\sin (60^\circ )( - \widehat i)$$
$$\varepsilon = {{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}( + \widehat i)$$
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