આકૃતિમાં દર્શાવ્યા મુજબ સમબાજુ ત્રિકોણ ABC | vedclass

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Here, $A C=B C=2 a$

$D$ and $E$ are the midpoints of $B C$ and $AC$.

$	herefore A E=E C=a$ and $B D=D C=a$

$\operatorname{In} \Delta A D C,(

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Vedclass · Answer

Here, $A C=B C=2 a$

$D$ and $E$ are the midpoints of $B C$ and $AC$.

$	herefore A E=E C=a$ and $B D=D C=a$

$\operatorname{In} \Delta A D C,(A D)^{2}=(A C)^{2}-(D C)^{2}$

$=(2 a)^{2}-(a)^{2}=4 a^{2}-a^{2}=3 a^{2}$

$A D=a \sqrt{3}$

Similarly, potential at point $D$ due to the given charge configuration is

$V_{D}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{B D}+\frac{q}{D C}+\frac{q}{A D}ight]$

$=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{a}+\frac{1}{a}+\frac{1}{\sqrt{3} a}ight]=\frac{q}{4 \pi \varepsilon_{0} a}\left[2+\frac{1}{\sqrt{3}}ight].........(i)$

Potential at point $E$ due to the given charge configuration is $V_{E}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A E}+\frac{q}{E C}+\frac{q}{B E}ight]$

$=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{a}+\frac{1}{a}+\frac{1}{a \sqrt{3}}ight]=\frac{q}{4 \pi \varepsilon_{0} a}\left[2+\frac{1}{\sqrt{3}}ight].........(ii)$

From the $(i)$ and $(ii)$, it is clear that

$V_{D}=V_{E}$

The work done in taking a charge $Q$ from $D$ to $E$ is

$W = Q({V_E} - {V_D}) = O$ $(\because \,{V_D} = {V_E})$

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