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The nucleus of a helium atom contains two protons that are separated by distance $3.0 \times {10^{ - 15}}m$ . The magnitude of the electrostatic force that each proton exerts on the other is:A. $20.6N$B. $25.6N$C. $15.6N$D. $12.6N$

Last updated date: 21st Jun 2024
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Hint: In order to find the electrostatic force between the protons, we must know the charge of a proton. Use the formula of electrostatic force between the protons. Put the values of charge, distance and the value of coulomb’s constant.

In electro statistics, two charged bodies exerts some force on each other. This force is known as electrostatic force. The magnitude of this electrostatic force is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Mathematically, this is given as:
${F_e} = \dfrac{{{q_p}{q_p}}}{{4\pi {\varepsilon _0}{r^2}}}$
Where ${F_e}$ is the electrostatic force on each particle
${q_p}$ is the charge of a proton with ${q_p} = 1.6 \times {10^{ - 19}}C$
$\dfrac{1}{{4\pi {\varepsilon _0}}}$ is coulomb’s constant having value of $9 \times {10^9}$
And $r$ is the distance between the protons.
Substituting the given values in the above equation, we can get the value of electrostatic force
$$\Rightarrow {F_e} = \dfrac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{(3 \times {{10}^{ - 15}})}^2}}}$$
$$\Rightarrow {F_e} = \dfrac{{23.04 \times 10}}{9}$$
$$\therefore {F_e} = 25.6N$$
Therefore, the electrostatic force between that one proton will exert on another is of the magnitude $$25.6N$$ .

Option B is the correct answer.

The charge on an electron is of the same magnitude as the charge on a proton. But it is written as ${q_e} = - 1.6 \times {10^{ - 19}}C$ with a negative sign. This negative sign indicates that the electron is negatively charged. If we observe carefully, we find that the formula of electrostatic force is similar to the formula of gravitational force.