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}} $
Three point charges $q, q$ and $-2 q$ are placed at the corners of an equilateral triangle of side '$L$'. Calculate work done by extemal force in moving all the charges far apart without acceleration
Derive the formula for the electric potential energy of system of two charges.
A problem of practical interest is to make a beam of electrons turn at $90^o$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 × 10^{-17}\ J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is $y × 10^5\ N/C$ . The value of $y$ is
Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are a distance $2\,L$ apart, $C$ is the midpoint between $A$ and $B.$ The work done in moving a charge $+Q$ along the semicircle $CRD$ is
An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be.