$A$ 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:

Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta$ with the $x$-axis,then the angle $\theta$ is

Two point charges $-10 \mu C$ and $+5 \mu C$ are situated on the $X$-axis at $x=0$ and $x=\sqrt{2} \ m$. The point along the $X$-axis where the electric field becomes zero is:

The distance between a proton and an electron in a hydrogen atom is ${10^{ - 10}} \ m$. Both have a charge of magnitude $1.6 \times {10^{ - 19}} \ C$. The magnitude of the electric field intensity produced at the position of the electron due to the proton is:

Two charges $Q$ are placed at two vertices of an equilateral triangle of side $a$. What is the electric field at the third vertex?

Two point charges $(+Q)$ and $(-2Q)$ are placed at distances $a$ and $2a$ respectively from the origin on the $X$-axis. At what distance from the origin is the electric field zero?