Charge $Q$ is given a displacement $\vec r = a\hat i + b\hat j$ in an electric field $\vec E = E_1\hat i + E_2\hat j$ . The work done is
$Q({E_1}a + {E_2}b)$
$Q\sqrt {{{({E_1}a)}^2} + {{({E_2}b)}^2}} $
$Q({E_1} + {E_2})\sqrt {{a^2} + {b^2}} $
$Q\sqrt {({E_1}^2 + {E_2}^2)} \sqrt {{a^2} + {b^2}} $
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........$m/s$ $(g = 10\, ms^{-2})$
In the following diagram the work done in moving a point charge from point $P$ to point $A$, $B$ and $C$ is respectively as $W_A$, $W_B$ and $W_C$ , then
An elementary particle of mass $m$ and charge $ + e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$What is the closest possible approach of the incident particle
In a region, electric field varies as $E = 2x^2 -4$ where $x$ is the distance in metre from origin along $x-$ axis. A positive charge of $1\,\mu C$ is released with minimum velocity from infinity for crossing the origin, then
A simple pendulum with a bob of mass $m = 1\ kg$ , charge $q = 5\mu C$ and string length $l = 1\ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 × 10^6\ V/m$ at its bottom most point $A$ , as shown in figure. It is given a speed $u$ such that the particle leave the circular path at its topmost point $C$ . Find the speed $u$ . (Take $g = 10\ m/s^2$ )