What is the magnitude of a point charge which produces an electric field of $2\, N/coulomb$ at a distance of $60\, cm$ $(1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N - {m^2}/{C^2})$
$8 \times {10^{ - 11}}\,C$
$2 \times {10^{ - 12}}\,C$
$3 \times {10^{ - 11}}\,C$
$6 \times {10^{ - 10}}\,C$
The intensity of the electric field required to keep a water drop of radius ${10^{ - 5}}\, cm$ just suspended in air when charged with one electron is approximately
Two charges $\pm 10\; \mu C$ are placed $5.0\; mm$ apart. Determine the electric field at $(a)$ a point $P$ on the axis of the dipole $15 cm$ away from its centre $O$ on the side of the positive charge, as shown in Figure $(a),$ and $(b)$ a point $Q , 15\; cm$ away from $O$ on a line passing through $O$ and normal to the axis of the dipole, as shown in Figure.
Two point charges $( + Q)$ and $( - 2Q)$ are fixed on the $X-$axis at positions $a$ and $2a$ from origin respectively. At what positions on the axis, the resultant electric field is zero
Explain electric field and also electric field by point charge.
An oil drop of radius $2\, mm$ with a density $3\, g$ $cm ^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5}\, V\, m ^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? (consider $\left. g =9.81\, m / s ^{2}\right)$