$A$ 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:

Two charges $+4e$ and $+e$ are at a distance $x$ apart. At what distance must a charge $q$ be placed from charge $+e$ so that it is in equilibrium?

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

$A$ negative charge is placed at the centre of a non-conducting sphere. The direction of the electric field at any point on the surface of the sphere is

Two charges $+5\,\mu C$ and $+10\,\mu C$ are placed $20\,cm$ apart. The net electric field at the midpoint between the two charges is

$A$ thin infinite sheet charge and an infinite line charge of respective charge densities $+\sigma$ and $+\lambda$ are placed parallel at $5 \ m$ distance from each other. Points $P$ and $Q$ are at $\frac{3}{\pi} \ m$ and $\frac{4}{\pi} \ m$ perpendicular distance from the line charge towards the sheet charge,respectively. $E_P$ and $E_Q$ are the magnitudes of the resultant electric field intensities at points $P$ and $Q$,respectively. If $\frac{E_P}{E_Q} = \frac{4}{a}$ for $2|\sigma| = |\lambda|$,then the value of $a$ is ...........