The Electric field is given by $\overrightarrow{E}=\frac{\overrightarrow{F}}{q_0}$, here the test charge '$q_0$' should be
a) Infinitesimally small and positive

b) Infinitesimally small and negative

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:

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

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:

Point charges $+q$ and $-q$ are located at the vertices of a square with diagonals $2l$ as shown in the figure. The magnitude of the electric field strength at a point, located symmetrically with respect to the vertices of the square at a distance $x$ from its center is

A thin disc of radius $b=2a$ has a concentric hole of radius $a$ in it (see figure). It carries uniform surface charge $\sigma$  on it. If the electric field on its axis at a height $\text{h}\left(\text{h}<<\text{a}\right)$ from its centre is given as $\text{C}\text{h}$ then the value of $C$ is