Two small conducting spheres of equal radius have charges $ + 10\,\mu C$ and $ - 20\,\mu C$ respectively and placed at a distance $R$ from each other experience force ${F_1}$. If they are brought in contact and separated to the same distance, they experience force ${F_2}$. The ratio of ${F_1}$ to ${F_2}$ is
$1:8$
$-8:1$
$1:2$
$-2:1$
If $g_E$ and $g_M$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio (electronic charge on the moon/electronic charge on the earth) to be
Write general equation of Coulombian force on ${q_1}$ by system of charges ${q_1},{q_2},.......,{q_n}$.
The electric field between the two spheres of a charged spherical condenser
The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $=6.7 \times 10^{-11} \,Nm ^2 / kg ^2$, mass of an electron $=9.1 \times 10^{-31} \,kg$, charge on an electron $=1.6 \times 10^{-19} C$ )
The ratio of coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is $2.4 \times 10^{39}$. The ratio of the proportionality constant, $K=\frac{1}{4 \pi \varepsilon_0}$ to the Gravitational constant $G$ is nearly (Given that the charge of the proton and electron each $=1.6 \times 10^{-19}\; C$, the mass of the electron $=9.11 \times 10^{-31}\; kg$, the mass of the proton $=1.67 \times 10^{-27}\,kg$ ):