The charge on \(500~\text{cc}\) of water due to protons will be: 

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are \(1.6~\mathring{A}\) apart is:
\(\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9~ \text{Nm}^2 \text{C}^{-2}\right)\)

Two small spheres each having the charge \(+Q\) are suspended by insulating threads of length \(L\) from a hook. If this arrangement is taken in space where there is no gravitational effect, then the angle between the two suspensions and the tension in each will be:

1. \(180^\circ,\) \(\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{(2 L )^{2}}\)

2. \(90^\circ,\) \(\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{(L )^{2}}\)

3. \(180^\circ,\) \(\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{2 L ^{2}}\)

4. \(180^\circ,\) \(\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{ L ^{2}}\) 

Two charges \(+2\) C and \(+6\) C are repelling each other with a force of \(12\) N. If each charge is given \(-2\) C of charge, then the value of the force will be:

Suppose the charge of a proton and an electron differ slightly. One of them is \(-e,\) the other is \((e+\Delta e).\) If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance \(d\) (much greater than atomic size) apart is zero, then \(\Delta e\) is of the order of?
(Given the mass of hydrogen ${\mathrm{}}_{}$\(m_h = 1.67\times 10^{-27}~\text{kg}\)$$)
1. \(10^{-23}~\text{C}\)
2. \(10^{-37}~\text{C}\)
3. \(10^{-47}~\text{C}\)
4. \(10^{-20}~\text{C}\)

Two positive ions, each carrying a charge \(q\), are separated by a distance \(d\). If \(F\) is the force of repulsion between the ions, the number of electrons missing from each ion will be:
(\(e\) is the charge on an electron)