$A$ uniformly charged rod of length $4\,cm$ and linear charge density $\lambda = 30\,\mu C/m$ is placed as shown in the figure. Calculate the $x-$ component of the electric field at point $P$.

$A$ positively charged ball hangs from a silk thread. We put a positive test charge $q_0$ at a point and measure $F/q_0$. Then, it can be predicted that the electric field strength $E$ 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: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

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?

Two point charges $-q$ and $+q/2$ are situated at the origin $(0, 0, 0)$ and at the point $(a, 0, 0)$ respectively. The point along the $X$-axis where the electric field vanishes is:

Three equal charges $+q$ each are placed at the three vertices of an equilateral triangle. The electric field at the centroid of the triangle is . . . . . . . ('$r$' is the length of the side of the triangle).