If the Coulombian force acting between two protons separated by a distance $r$ is $F$, what would be the force acting between two alpha particles separated by a distance $2r$?

Assume that you have $10^{23}$ carbon atoms. All the nuclei are placed at the North Pole of the Earth and all the electrons are placed at the South Pole of the Earth. (Radius of Earth = $6400 \ km$). The force between the charges is approximately:

Three charges $+5 q, Q$ and $-2 q$ are kept along a straight line in the same order such that $+5 q$ and $-2 q$ charges are at a distance of $\frac{2 r}{3}$ and $\frac{r}{3}$ from the charge $Q$ respectively. If the net force on the charge $-2 q$ is zero,then $Q$ is

Consider the following charged suspended ball system. If $\alpha < \beta$,then out of the following statements,which can be true at equilibrium? :-
$(a) Q_1 > Q_2, m_1 < m_2$
$(b) Q_1 > Q_2, m_1 > m_2$
$(c) Q_1 < Q_2, m_1 = m_2$
$(d) Q_1 < Q_2, m_1 > m_2$

Two point charges $+10 \mu C$ and $+4 \mu C$ are placed $10 \ cm$ apart in air. The work required to be done to bring them $2 \ cm$ closer is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right)$. (in $J$)

Two positive point charges are separated by a distance of $4 \ m$ in air. If the sum of the two charges is $36 \mu C$ and the electrostatic force between them is $0.18 \ N$,then the bigger charge is (in $\mu C$)