When electric field is applied to the electrons in a conductor it starts

A charged particle (mass $\mathrm{m}$ and charge $\mathrm{q}$) moves along $\mathrm{X}$ axis with velocity ${\mathrm{V}}_0$. When it passes through the origin it enters a region having uniform electric field $\overrightarrow{\mathrm{E}}=-\mathrm{E}\hat{\mathrm{j}}$ which extends upto $\mathrm{x}=\mathrm{d}$. Equation of path of electron in the region $\mathrm{x}>\mathrm{d}$ is:

Charges ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ are at points $\mathrm{A}$ and $\mathrm{B}$ of a right-angled triangle $OAB$. The resultant electric field at point $\mathrm{O}$ is perpendicular to the hypotenuse, then ${\mathrm{Q}}_1/{\mathrm{Q}}_2$ is proportional to:

A uniform electric field, $\overrightarrow{E}=-400\sqrt{3} \hat{\mathrm{j}} \mathrm{N} {\mathrm{C}}^{-1}$ is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2\sqrt{10}\times {10}^6 \mathrm{m} {\mathrm{s}}^{-1}$. This particle is aimed to hit a target $T$, which is $5 \mathrm{m}$ away from its entry point into the field as shown schematically in the figure. Take $\frac{q}{m}={10}^{10} \mathrm{C} {\mathrm{kg}}^{-1}$. Then

Three charged particles $\text{A}, \text{B}$ and $\text{C}$ with charges $-4q,2q$ and $-2q$ are present on the circumference of a circle of radius $d.$ The charged particles $\text{A},\text{C}$ and centre $\text{O}$ of the circle formed an equilateral triangle as shown in the figure. The electric field at the point $\text{O}$ is