$A$ 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:

$A$ 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 ...........

Two point unlike charges having magnitudes $+16 \mu C$ and $-9 \mu C$ are separated by a distance of $10 \ cm$ in air. The resultant electric field will be zero at a distance of . . . . . . from the $-9 \mu C$ charge. (in $cm$)

Four point charges $-q, +q, +q$ and $-q$ are placed on the $y$-axis at $y = -2d, y = -d, y = +d$ and $y = +2d$,respectively. The magnitude of the electric field $E$ at a point on the $x$-axis at $x = D$,with $D >> d$,will vary as:

For the given figure,the direction of the electric field at point $A$ will be:

Four charges,each of magnitude $q$ coulomb,are placed at points $(-1,0,0), (1,0,0), (0,-1,0)$ and $(0,1,0)$ in the $xy$-plane. The distances along the axes are measured in metres. The magnitude of the electric field at the point $(0,0,1)$ on the $Z$-axis is