NEET-XII-Physics
04: The Forces
- #10Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.Ans : Charge of the proton, q = `` 1.6\times {10}^{-19}\,\mathrm{\,C \,}``
Mass of the proton = `` 1.67\times {10}^{-27}\,\mathrm{\,kg \,}``
Let the distance between two protons be r.
Coulomb force (electric force) between the protons is given by
`` {f}_{e}=\frac{1}{4\,\mathrm{\,\pi \,}{\in }_{0}}\times \frac{{q}^{2}}{{r}^{2}}``
`` =\frac{9\times {10}^{9}\times (1.6{)}^{2}\times {10}^{-38}}{{r}^{2}}``
Gravitational force between the protons is given by
`` {f}_{g}=\frac{\,\mathrm{\,G \,}{m}^{2}}{{r}^{2}}``
`` =\frac{6.67\times {10}^{-11}\times (1.67\times {10}^{-27}{)}^{2}}{{r}^{2}}``
On dividing `` {f}_{e}\,\mathrm{\,by \,}{f}_{g}``, we get:
`` \frac{{f}_{e}}{{f}_{g}}=\frac{1}{4\,\mathrm{\,\pi \,}{\in }_{0}}\times \frac{{q}^{2}}{{r}^{2}}\times \frac{{r}^{2}}{\,\mathrm{\,G \,}{m}^{2}}``
`` =\frac{9\times {10}^{9}\times 1.6\times 1.6\times {10}^{-38}}{6.67\times {10}^{-11}\times 1.67\times 1.67\times {10}^{-54}}``
`` =\frac{9\times (1.6{)}^{2}\times {10}^{-29}}{6.67\times (1.67{)}^{2}\times {10}^{-65}}``
`` =1.24\times {10}^{36}``