Three charges,each +q,are placed at the corne | vedclass

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(D) Given that $AC = BC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively.Therefore,$AE = EC = a$ and $BD = DC = a$.In $\Delta ADC$,b

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(D) Given that $AC = BC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively.Therefore,$AE = EC = a$ and $BD = DC = a$.In $\Delta ADC$,by the Pythagorean theorem,$(AD)^2 = (AC)^2 - (DC)^2 = (2a)^2 - (a)^2 = 4a^2 - a^2 = 3a^2$,so $AD = a\sqrt{3}$.Similarly,in $\Delta BEC$,$(BE)^2 = (BC)^2 - (EC)^2 = (2a)^2 - (a)^2 = 3a^2$,so $BE = a\sqrt{3}$.The electric potential at point $D$ due to the charges at $A, B,$ and $C$ is:$V_D = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{BD} + \frac{q}{DC} + \frac{q}{AD} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{a} + \frac{1}{a} + \frac{1}{a\sqrt{3}} \right] = \frac{q}{4\pi\varepsilon_0 a} \left[ 2 + \frac{1}{\sqrt{3}} \right]$.The electric potential at point $E$ due to the charges at $A, B,$ and $C$ is:$V_E = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AE} + \frac{q}{EC} + \frac{q}{BE} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{a} + \frac{1}{a} + \frac{1}{a\sqrt{3}} \right] = \frac{q}{4\pi\varepsilon_0 a} \left[ 2 + \frac{1}{\sqrt{3}} \right]$.Since $V_D = V_E$,the work done $W = Q(V_E - V_D) = 0$.

Three charges,each $+q$,are placed at the corners of an isosceles triangle $ABC$ with sides $BC = AC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$,respectively. The work done in taking a charge $Q$ from $D$ to $E$ is

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