Three positive charges $q$ are placed at the vertices of an equilateral triangle. How do their electric field lines appear?

$A$ table tennis ball,which has been covered with conducting paint,is suspended by a silk thread so that it hangs between two plates. One plate is earthed,and the other is connected to a high-voltage generator. This ball:

Five balls numbered $1$ to $5$ are suspended using separate threads. Pairs $(1,2), (2,4)$ and $(4,1)$ show electrostatic attraction while pairs $(2,3)$ and $(4,5)$ show repulsion. Therefore,ball $1$ must be:

An infinite plane sheet of charge having uniform surface charge density $+\sigma_s \text{ C/m}^2$ is placed on the $x-y$ plane. Another infinitely long line charge having uniform linear charge density $+\lambda_e \text{ C/m}$ is placed at the $z=4 \text{ m}$ plane and is parallel to the $y$-axis. If the magnitude values satisfy $|\sigma_s| = 2|\lambda_e|$,then at the point $(0, 0, 2)$,the ratio of the magnitudes of the electric field values due to the sheet charge to that of the line charge is $\pi \sqrt{n} : 1$. The value of $n$ is:

$A$ glass rod rubbed with silk is used to charge a gold leaf electroscope and the leaves are observed to diverge. The electroscope thus charged is exposed to $X$-rays for a short period. Then

The earth's surface has a negative surface charge density of $10^{-9} \; C\;m^{-2}$. The potential difference of $400 \; kV$ between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only $1800 \; A$ over the entire globe. If there were no mechanism of sustaining the atmospheric electric field,how much time (roughly) would be required to neutralize the earth's surface (in $; s$)? (This never happens in practice because there is a mechanism to replenish electric charges,namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth $= 6.37 \times 10^{6} \; m$.)