दो वैध्युत आवेशों के बीच स्थिर वैध्युत बल के लिए कूलॉम नियम तथा दो स्थिर बिंद्रु द्रव्यमानों के बीच गुर्त्वाकर्षण बल के लिए न्यूटन का नियम दोनों में ही बल आवेशों/द्रव्यमानों के बीच की दूरी के वर्ग के व्युत्क्रमानुपाती होता है। $(a)$ इन दोनों बलों के परिमाण ज्ञात करके इनकी प्रबलताओं की तुलना की जाए $(i)$ एक इलेक्ट्रॉन तथा एक प्रोटॉन के लिए,
$(ii)$ दो प्रोटनों के लिए। $(b)$ इलेक्ट्रॉन तथा प्रोटॉन में पारस्परिक आकर्षण के वैध्युत बल के कारण इलेक्ट्रॉन तथा प्रोटॉन के त्वरण आकलित कीजिए जबकि इनके बीच की दूरी $1 \dot{ A }\left(=10^{-10} \,m \right)$ है। $\left(m_{p}=1.67 \times 10^{-27} \,K , m_{e}=9.11 \times 10^{-31} \,kg \right)$

$(i)$ The electric force between an electron and a proton at a distance $r$ apart is:

$F_{e}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$

where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is:

$F_{c}=-G \frac{m_{p} m_{e}}{r^{2}}$

where $m_{p}$ and $m_{e}$ are the masses of a proton and an electron respectively.

$\left|\frac{F_{e}}{F_{G}}\right|=\frac{e^{2}}{4 \pi \varepsilon_{0} G m_{p} m_{e}}=2.4 \times 10^{39}$

$(ii)$ On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance $r$ apart is:

$\left|\frac{F_{e}}{F_{G}}\right|=\frac{e^{2}}{4 \pi \varepsilon_{0} G m_{p} m_{p}}=1.3 \times 10^{36}$

However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values of these forces between two protons inside a nucleus (distance between two protons is $ \sim {10^{ - 15}}\,m$ inside a nucleus) are $F_{e} \sim 230\, N$ whereas, ${F_G} \sim 1.9 \times {10^{ - 34}}\,N$

The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

$(b)$ The electric force $F$ exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton: however. the masses of an electron and a proton are different. Thus, the magnitude of force is

$\left| F \right| = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{{r^2}}} = 8.987 \times {10^9}N\frac{{N{m^2}}}{{{C^2}}}\frac{{{{\left( {1.6 \times {{10}^{ - 19}}\,C} \right)}^2}}}{{{{\left( {{{10}^{ - 10}}\,m} \right)}^2}}}$

$=2.3 \times 10^{-8} \,N$

Using Newton's second law of motion. $F=m a$, the acceleration that an electron will undergo is

$a=2.3 \times 10^{-8}\, N / 9.11 \times 10^{-31}\, kg $$=2.5 \times 10^{22} \,m / s ^{2}$

Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton.

The value for acceleration of the proton is $2.3 \times 10^{-8} \,N / 1.67 \times 10^{-27} \,kg $$=1.4 \times 10^{19}\, m / s ^{2}$