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:

Four point positive charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame of side $L$ as shown in the figure. The plane of the frame is perpendicular to the $Z$-axis. If a negative point charge $(-q)$ is placed at a small distance $z$ away from the center of the frame along the $Z$-axis $(z < < L)$, then:

Two charges $+Q$ and $-2 Q$ are located at points $A$ and $B$ on a horizontal line as shown below. The electric field is zero at a point which is located at a finite distance:

$A$ charged particle having some mass is resting in equilibrium at a height $H$ above the centre of a uniformly charged non-conducting horizontal ring of radius $R$. The force of gravity acts downwards. The equilibrium of the particle will be stable if:

An infinite number of electric charges, each equal to $5 \text{ nC}$ (magnitude), are placed along the $X$-axis at $x = 1 \text{ cm}, x = 2 \text{ cm}, x = 4 \text{ cm}, x = 8 \text{ cm}, \dots$ and so on. In this setup, if the consecutive charges have opposite signs, then the electric field in $\text{N/C}$ at $x = 0$ is: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

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