The value of electric permittivity of free space is
$9 \times {10^9}\,N{C^2}/{m^2}$
$8.85 \times {10^{ - 12}}\,N{m^2}/{C^2}sec$
$8.85 \times {10^{ - 12}}\,{C^2}/N{m^2}$
$9 \times {10^9}\,{C^2}/N{m^2}$
Two positive point charges of unequal magnitude are placed at a certain distance apart. A small positive test charge is placed at null point, then
An electric field due to a positively charged long straight wire at a distance $r$ from it is proportional to $r^{-1}$ in magnitude. Two electrons are orbiting such a long straight wire in circular orbits of radii $1 A$ and $2 A$. The ratio of their respective time periods is
Two point charges $A$ and $B$, having charges $+Q$ and $- Q$ respectively, are placed at certain distance apart and force acting between them is $\mathrm{F}$. If $25 \%$ charge of $A$ is transferred to $B$, then force between the charges becomes
Four charges are placed at the circumference of the dial of a clock as shown in figure. If the clock has only hour hand, then the resultant force on a positive charge $q_0$ placed at the centre, points in the direction which shows the time as
$5$ charges each of magnitude $10^{-5} \,C$ and mass $1 \,kg$ are placed (fixed) symmetrically about a movable central charge of magnitude $5 \times 10^{-5} \,C$ and mass $0.5 \,kg$ as shown in the figure given below. The charge at $P_1$ is removed. The acceleration of the central charge is [Given, $\left.O P_1=O P_2=O P_3=O P_4=O P_5=1 m , \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right]$