Question

Point charges q, -q, 2Q and Q are placed at the corners of A,B,C,D of a square of side 2b. If the field at the midpoint of CD is zero, then $\dfrac{q}{Q}$is
a) 1
b) 2
c) $\dfrac{2\sqrt{2}}{5}$
d) $\dfrac{5\sqrt{5}}{2}$

Answer 1

Hint: First draw the diagram to see the distribution of charges along the square. It is given that the electric field at the midpoint on the side of the square CD is zero. Hence we will use the expression of electric field at a point due to a charge i.e. $E=\dfrac{kq}{{{r}^{2}}}$ (where k is the permittivity of free space, q is the charge due to which field is generated and r is the distance at which the value of electric field due to charge q is to be calculated) and add the field due to all the charges vector ally at the midpoint of the side CD as zero and obtained the required ratio.

Complete step-by-step answer:
To begin with let us first draw the configuration of the charges arranged.

In the above diagram, the point M is the midpoint of side DC. The angle $\angle BMC$ is half the angle between the vectors ${{E}_{(q)}}$ and ${{E}_{(-q)}}$. The field due to q is denoted by ${{E}_{(q)}}$ and is directed along AM, the field due to –q is denoted by ${{E}_{(-q)}}$ and is directed along MB, the field due to Q is denoted by ${{E}_{(Q)}}$ and is directed along MC, the field due to 2Q is denoted by ${{E}_{(2Q)}}$ and is directed along MD. The size of the side AM is $\sqrt{\text{5}}\text{b (by pythagoras theorem})$. The electric fields ${{E}_{(Q)}}$ and ${{E}_{(2Q)}}$ are directed in opposite directions. Hence their net value at point M can be mathematically written as,
$\begin{align}
  & {{E}_{1}}={{E}_{(Q)}}+(-){{E}_{(2Q)}} \\
 & {{E}_{1}}=\dfrac{kQ}{{{b}^{2}}}-\dfrac{k2Q}{{{b}^{2}}}=-\dfrac{kQ}{{{b}^{2}}} \\
\end{align}$
The minus sign indicates that the net electric field is along MD.
Let us use now the laws of vector addition to find the value of the electric field due to q and –q at point M.
${{E}_{2}}=\sqrt{{{\left( {{E}_{(q)}} \right)}^{2}}+{{\left( {{E}_{(-q)}} \right)}^{2}}+2{{E}_{(q)}}{{E}_{(-q)}}\operatorname{cos}2\angle BMC}$
Now lets us first solve for $\operatorname{cos}2\angle BMC$ separately.
By trigonometric definition, $\operatorname{cos}2\theta ={{\operatorname{cos}}^{2}}\theta -{{\operatorname{sin}}^{2}}\theta $ ,hence $\operatorname{cos}2\angle BMC$becomes,
$\begin{align}
  & \operatorname{cos}2\angle BMC={{\operatorname{cos}}^{2}}\angle BMC-{{\operatorname{sin}}^{2}}\angle \text{BMC}\text{. If we see in the figure,} \\
 & \text{cos}\angle BMC=\dfrac{b}{\sqrt{5}b}\text{ and }\operatorname{sin}\angle \text{BMC=}\dfrac{2b}{\sqrt{5}b},\text{hence the above equation becomes,} \\
 & \operatorname{cos}2\angle BMC={{\operatorname{cos}}^{2}}\angle BMC-{{\operatorname{sin}}^{2}}\angle \text{BMC} \\
 & \operatorname{cos}2\angle BMC={{\left( \dfrac{b}{\sqrt{5}b} \right)}^{2}}-{{\left( \dfrac{2b}{\sqrt{5}b} \right)}^{2}}=\dfrac{{{b}^{2}}}{5{{b}^{2}}}(1-4)=-\dfrac{3}{5} \\
\end{align}$
Now let us substitute this value in the above equation.
$\begin{align}
  & {{E}_{2}}=\sqrt{{{\left( \dfrac{kq}{{{(\sqrt{5}b)}^{2}}} \right)}^{2}}+{{\left( \dfrac{kq}{{{(\sqrt{5}b)}^{2}}} \right)}^{2}}+2\dfrac{kq}{{{(\sqrt{5}b)}^{2}}}\dfrac{kq}{{{(\sqrt{5}b)}^{2}}}\times -\dfrac{3}{5}} \\
 & {{E}_{2}}=\sqrt{{{\left( \dfrac{kq}{{{(\sqrt{5}b)}^{2}}} \right)}^{2}}+{{\left( \dfrac{kq}{{{(\sqrt{5}b)}^{2}}} \right)}^{2}}+2{{\left( \dfrac{kq}{{{(\sqrt{5}b)}^{2}}} \right)}^{2}}\times -\dfrac{3}{5}} \\
 & {{E}_{2}}=\dfrac{kq}{{{(\sqrt{5}b)}^{2}}}\sqrt{2-\dfrac{6}{5}}=\dfrac{kq}{{{(\sqrt{5}b)}^{2}}}\sqrt{\dfrac{10-6}{5}}=\dfrac{2kq}{\sqrt{5}{{(\sqrt{5}b)}^{2}}} \\
\end{align}$
It is given in the question that the net field at M is zero. Hence,
$\sqrt{{{E}_{1}}^{2}+{{E}_{2}}^{2}}=0,\text{which implies }{{E}_{1}}={{E}_{2}}$ .Therefore after equating both of them we get,
$\begin{align}
  & \dfrac{kQ}{{{b}^{2}}}=\dfrac{2kq}{\sqrt{5}{{(\sqrt{5}b)}^{2}}} \\
 & Q=\dfrac{2q}{5\sqrt{5}} \\
 & \dfrac{q}{Q}=\dfrac{5\sqrt{5}}{2} \\
\end{align}$

So, the correct answer is “Option d)”.

Note: The minus sign next to the value of the electric field indicates the direction of the field. Hence while calculating the net electric field the minus sign of the electric field should be ignored as we only calculate its magnitude. The electric field due to a positive charge is away from itself whereas an electric field due to negative charge is directed towards itself.