Question

Point charges $q,-q,2Q$ and $Q$ are placed in order at the corners $A,B,C,D$ of a square of side $2b$. If the field at the midpoint $CD$ is zero, then $\frac{q}{Q}$ is:

• A. $1$
• B. $2$
• C. $\frac{2\sqrt{2}}{5}$
• D. $\frac{5\sqrt{5}}{2}$

Answer 1

Answer: (D)

$\frac{5\sqrt{5}}{2}$

Point charges $q$, $-q$, $2Q$ and $Q$ are placed in order to the corners. Here, four point charges $+q$, $2Q$, $-q$, $Q$ are given along with diagonal side.

$\sqrt{{\left(2b\right)}^2+{\left(2b\right)}^2}$

$=\sqrt{{4b}^2+{4b}^2}$

$=\sqrt{{8b}^2}$

$=b\sqrt{4\times 2}$

$=2b\sqrt{2}$

Now, the electric field at the intersection point of the diagonals of the square. Hence, the net electric field.

$E=K\left(\frac{q+2Q-q+Q}{r^2}\right)$

$=\frac{3Q}{r^2}K$

$=9\times {10}^9\times \frac{3\times Q}{2b\sqrt{2}}$

$=\frac{5\sqrt{5}}{2}$