Question

Consider a parallelogram $ABCD$, with angle $\angle B={120}^o$. A charge $+Q$ placed at the corner $A$ produces field $E$ and potential $V$ at corner $D$. If we now add charges $-2Q$ and $+Q$ at corners $B$ and $C$ respectively, the magnitude of field and potential at $D$ will become respectively :

• A. $\frac{\sqrt{3}}{2}E,V$
• B. $\frac{\sqrt{3}}{2}E,0$
• C. $E,0$
• D. $\frac{E}{\sqrt{2}},\frac{V}{\sqrt{2}}$

Answer 1

The correct option is C $E,0$

Here distance is same for every charge from given point $D$ which is $a$ as shown in the fig.

$\therefore V_{-2Q}=-2V$ and $V_{+Q}=V$

$\therefore$ Net Potential at point $D$ is given by,

$V_D=V_{+Q}+V_{-2Q}+V_{+Q}=V-2V+V=0$

As shown in the fig net electric field due to all charges at point $D$ is given by,

$E_D=\sqrt{(2Ecos{30}^o-Ecos{30}^o{)}^2+(2Esin{30}^o+Esin{30}^o-E{)}^2}$

$E_D=\sqrt{(Ecos{30}^o{)}^2+(Esin{30}^o{)}^2}=\sqrt{E^2}$

$\therefore E_D=E$ and $V_D=0$