Two charges each equal to eta q({eta ^{ - 1}} | vedclass

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(c) ${E_1} = \frac{{\eta \,q}}{{4\pi {\varepsilon _0}{a^2}}},\,{E_2} = \frac{{\eta \,q}}{{4\pi {\varepsilon _0}{a^2}}}.$ Therefore $E = {\vec E_1} + {

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${E_3} > {E_0}$

Vedclass · Answer

(c) ${E_1} = \frac{{\eta \,q}}{{4\pi {\varepsilon _0}{a^2}}},\,{E_2} = \frac{{\eta \,q}}{{4\pi {\varepsilon _0}{a^2}}}.$ Therefore $E = {\vec E_1} + {\vec E_2}$ $ = \sqrt {E_1^2 + E_2^2 + 2{E_1}{E_2}\cos {{60}^o}} = \frac{{\sqrt 3 \eta \,q}}{{4\pi {\varepsilon _0}{a^2}}}.$ Since ${\eta ^{ - 1}} < \sqrt 3 ,\,1 < \sqrt 3 \eta ,\,\sqrt 3 \eta > 1.$ $==>$ $\frac{{\sqrt 3 \eta q}}{{4\pi {\varepsilon _0}{a^2}}} > \frac{q}{{4\pi {\varepsilon _0}{a^2}}}$ $==>$ ${E_3} > {E_0}\,\left( {{E_0} = \frac{q}{{4\pi {\varepsilon _0}{a^2}}}} \right)$.

Two charges each equal to $\eta q({\eta ^{ - 1}} < \sqrt 3 )$ are placed at the corners of an equilateral triangle of side $a$. The electric field at the third corner is ${E_3}$ where $({E_0} = q/4\pi {\varepsilon _0}{a^2})$

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