A small sphere of mass m = 0.5, kg carrying a positive charge q = 110 μ C is connec

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2. Electric Potential and Capacitance

hard

A small sphere of mass $m =\ 0.5\, kg$ carrying a positive charge $q = 110\ \mu C$ is connected with a light, flexible and inextensible string of length $r = 60 \ cm$ and whirled in a vertical circle. If a vertically upwards electric field of strength $E = 10^5 NC^{-1}$ exists in the space, The minimum velocity of sphere required at highest point so that it may just complete the circle........$m/s$ $(g = 10\, ms^{-2})$

$8$

$7$

$6$

$9$

Solution

$\mathrm{Fg}=5 \mathrm{\,N}, \quad \mathrm{Fe}=11 \mathrm{\,N}$

The minimum possible velocity at lowest point can be $\sqrt{7.2} \mathrm{\,m} / \mathrm{s} .$ The corresponding velocity at the highest point will also be minimum

${\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}=-m g(2 r)+q E_{0}(2 r)} $

${v_{s}=6 \,m / s}$

Standard 12

Physics

If $4 \times {10^{20}}eV$ energy is required to move a charge of $0.25$ coulomb between two points. Then what will be the potential difference between them……$V$

easy

A ball of mass $1\, g$ and charge ${10^{ – 8}}\,C$ moves from a point $A$. where potential is $600\, volt$ to the point $B$ where potential is zero. Velocity of the ball at the point $B$ is $20\, cm/s$. The velocity of the ball at the point $A$ will be

medium

If $OP = 1\,\,cm$ and $OS = 2\,\, cm$, work done by electric field in shifting a point charge $\frac {4\sqrt 2}{27}\,\, μC$ from point $P$ to $S$ in given figure is

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Electrostatic potential energy of given system will be