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: ($m_p=1.67\times10^{-27}~\text{kg},~m_e=9.11\times10^{-31}~\text{kg}$)