Two point charges of $20\,\mu C$ and $80\,\mu C$ are $10\,cm$ apart. Where will the electric field strength be zero on the line joining the charges from the $20\,\mu C$ charge? (in meters)

The unit of electric field intensity is . . . . . . .

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: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

Three charges $2q, -q$ and $-q$ are located at the vertices of an equilateral triangle. At the center of the triangle,

$A$ unit negative charge with mass $M$ resides at the mid-point of the straight line of length $2a$ connecting two fixed charges of magnitude $+Q$ each. If it is given a very small displacement $x$ $(x \ll a)$ in a direction perpendicular to the straight line, it will:

$A$ semi-infinite non-conducting rod lies along the $+x$-axis with its left end at the origin. The rod has a uniform linear charge density $\lambda$. The magnitude of the electric field $|\vec{E}|$ at a point on the $y$-axis at a distance $L$ from the origin will be: