Two point charges $(+Q)$ and $(-2Q)$ are fixed on the $X$-axis at positions $a$ and $2a$ from the origin,respectively. At what position on the axis is the resultant electric field zero?

$A$ point charge $+q$ is placed at the origin. $A$ second point charge $+9q$ is placed at $(d, 0, 0)$ in a Cartesian coordinate system. The point in between them where the electric field vanishes is:

The electric field due to a charge at a distance of $3\, m$ from it is $500\, N/C$. The magnitude of the charge is $.......\,\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N \cdot m^2}}{{C^2}}} \right]$

$A$ thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at the centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is . . . . . . $\text{nC}$. ($\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)

If the net electric field at point $P$ along the $Y$-axis is zero,then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$,where $x = . . . . . .$

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \ N/C$ normally. $A$ charged particle of mass $2 \ g$ is suspended by a silk thread of length $20 \ cm$ and remains at a distance of $10 \ cm$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} \ \mu C$ where $x=$ . . . . . . . (Use $g=10 \ m/s^2$)