The acceleration of an electron in an electric field of magnitude $50\, V/cm$, if $e/m$ value of the electron is $1.76 \times {10^{11}}\,C/kg$, is
$8.8 \times {10^{14}}\,m/sec^2$
$6.2 \times {10^{13}}\,m/sec^2$
$5.4 \times {10^{12}}\,m/sec^2$
Zero
Two point charges of $20\,\mu \,C$ and $80\,\mu \,C$ are $10\,cm$ apart. Where will the electric field strength be zero on the line joining the charges from $20\,\mu \,C$ charge......$m$
What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,newton/coulomb$ $[1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N{m^2}/{C^2}]$
The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\, N/C$ just above its surface is
The direction $(\theta ) $ of $\vec E$ at point $P$ due to uniformly charged finite rod will be
Write equation of electric field by point charge. How does it depend on distance ?