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(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Charges $Q$ are placed at $B$ and $C$.The electric field at vertex $A$ due to cha

Vedclass · Accepted Answer

$\frac{\sqrt{3} Q}{4\pi \varepsilon_0 a^2}$

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(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Charges $Q$ are placed at $B$ and $C$.The electric field at vertex $A$ due to charge at $B$ is $E_B = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{a^2}$ directed along $BA$.The electric field at vertex $A$ due to charge at $C$ is $E_C = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{a^2}$ directed along $CA$.The angle between $E_B$ and $E_C$ is $60^\circ$.The resultant electric field $E$ is given by the vector sum:$E = \sqrt{E_B^2 + E_C^2 + 2E_B E_C \cos 60^\circ}$Since $E_B = E_C = E_0 = \frac{Q}{4\pi \varepsilon_0 a^2}$,we have:$E = \sqrt{E_0^2 + E_0^2 + 2E_0^2 \cdot \frac{1}{2}} = \sqrt{3E_0^2} = \sqrt{3} E_0$$E = \frac{\sqrt{3} Q}{4\pi \varepsilon_0 a^2}$

Two charges $Q$ are placed at two vertices of an equilateral triangle of side $a$. What is the electric field at the third vertex?

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