The ratio of the magnitude of electric force to the magnitude of gravitational force for an electron and a proton will be:
(\(m_p=1.67\times10^{-27}~\text{kg}\), \(m_e=9.11\times10^{-31}~\text{kg}\))
1. \(2.4\times10^{39}\)
2. \(2.6\times10^{36}\)
3. \(1.4\times10^{36}\)
4. \(1.6\times10^{39}\)

Consider three charges \(q_1,~q_2,~q_3\) each equal to \(q\) at the vertices of an equilateral triangle of side \(l.\) What is the force on a charge \(Q\) (with the same sign as \(q\)) placed at the centroid of the triangle, as shown in the figure?

     
1. \(\dfrac{3}{4\pi \epsilon _{0}} \dfrac{Qq}{l^2}\)
2. \(\dfrac{9}{4\pi \epsilon _{0}} \dfrac{Qq}{l^2}\)
3. zero
4. \(\dfrac{6}{4\pi \epsilon _{0}} \dfrac{Qq}{l^2}\)

Consider the charges \(q,~q,\) and \(-q\) placed at the vertices of an equilateral triangle, as shown in the figure. Then the sum of the forces on the three charges is:

    

1. \(\frac{1}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}\)
2. zero
3. \(\frac{2}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}\)
4. \(\frac{3}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}\)

The accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (=${10}^{-10} m$) apart are respectively: (\(m_p=1.67\times10^{-27}~\text{kg},~m_e=9.11\times10^{-31}~\text{kg}\))