An electric field is given by $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$. $A$ charge $Q$ is moved by a displacement vector $\vec{r} = a \hat{i} + b \hat{j}$. The work done is:

$A$ point charge of $0.009 \ \mu C$ is placed at the origin. Calculate the electric field intensity due to this point charge at the point $(\sqrt{2}, \sqrt{7}, 0)$.

What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,N/C$ $[1/4\pi \varepsilon_0 = 9 \times 10^9\,N\cdot m^2/C^2]$?

$A$ deuteron and an $\alpha$-particle are placed $1\,\mathring{A}$ apart in air. The magnitude of the intensity of the electric field due to the deuteron at the position of the $\alpha$-particle 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: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

$A$ deuteron and an $\alpha$-particle are separated by a distance of $1\,\mathring{A}$ in air. The magnitude of the electric field due to the deuteron at the position of the $\alpha$-particle is: