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Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields
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Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges/masses. Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are $1 \mathring{A} =(10^{-10}\;m) $ apart? $(mp = 1.67 × 10^{–27}\; kg, m_e = 9.11 × 10^{–31}\; kg)$

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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
$|F| = \large\frac{1}{4 \pi \in_0} \frac{e^2}{r^2}$$=8.987 \times 10^9 \;Nm^2/C^2 \times (1.6 \times 10^{-19} C)^2 / (10^{-10}m)^2$
$\qquad =2.3 \times 10^{-8}\;N$
Using Newton’s second law of motion,$ F = ma,$ the acceleration that an electron will undergo is
$a= 2.3 \times 10^{-8} N /9.11 \times 10^{-31} \;kg$
$\quad= 2.5 \times 10^{22}m/s^2$
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$
$\qquad =1.4 \times 10^{19}m/s^2$
Hence B is the correct answer.
answered May 29, 2014 by meena.p
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