Question 371: Electrostatics & Capacitance

Question

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 ?

Accepted Answer

$$\varepsilon  = {{2k\lambda } \over R}\sin \left( {{	heta  \over 2}} ight)( - \widehat i)$$
$$\lambda  = \left( {{{ - Q} \over {R	heta }}} ight) = \left( {{{ - Q} \over {R.{{2\pi } \over 3}}}} ight)$$
$$\lambda  = {{ - 3Q} \over {2\pi R}}$$
$$\varepsilon  = {{2k} \over R}.{{ - 3Q} \over {2\pi R}}.\sin (60^\circ )( - \widehat i)$$
$$\varepsilon  = {{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}( + \widehat i)$$

Answer

$${{3\sqrt 3 Q} \over {8\pi {\varepsilon _0}{R^2}}}(\widehat i)$$

Answer

$${{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}(\widehat i)$$

Answer

$${{3\sqrt 3 Q} \over {16{\pi ^2}{\varepsilon _0}{R^2}}}(\widehat i)$$

Answer

$${{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}( - \widehat i)$$

Electrostatics & Capacitance

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 ?

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