A particle of mass $m$ and charge $q$ is kept at the top of a fixed frictionless sphere. $A$ uniform horizontal electric field $E$ is switched on. The particle looses contact with the sphere, when the line joining the center of the sphere and the particle makes an angle $45^o$ with the vertical. The ratio $\frac{qE}{mg}$ is :-
$\frac{3}{{3 + 2\sqrt 2 }}$
$\frac{{3 + 2\sqrt 2 }}{3}$
$\frac{3}{{3 - 2\sqrt 2 }}$
$\frac{{3 - 2\sqrt 2 }}{3}$
Two equal charges $q$ are placed at a distance of $2a$ and a third charge $ - 2q$ is placed at the midpoint. The potential energy of the system is
Three point charges $Q, 4Q $ and $16Q $ are placed on a straight line $9$ $cm$ long. Charges are placed in such a way that the system has minimum potential energy. Then
A bullet of mass $2\, gm$ is having a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$
In free space, a particle $A$ of charge $1\,\mu C$ is held fixed at a point $P.$ Another particle $B$ of the same charge and mass $4\,\mu g$ is kept at a distance of $1\,mm$ from $P$. If $B$ is released, then its velocity at a distance of $9\,mm$ from $P$ is [ Take $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N{m^2}{C^{ - 2}}$ ]
The figure shows a family of parallel equipotential surfaces and four paths along which an electron is made to move from one surface to another as shown in the figur
$(I)$ What is the direction of the electric field ?
$(II)$ Rank the paths according to magnitude of work done, greatest first