Calculate force of repulsion between two protons when they are separated with 4.0 10 15 m

Text Solution

Solution : KEY IDEAS <br> The protons can be treated as charged particles, so the magnitude of the electrostatic force on one from the other is given by Coulomb.s law. <br> Calculation: Table 22-1 tells us that the charge of a proton is +e. Thus, Eq. 22-4 gives us <br> `F=(1)/(4pi epsilon_(0)) (e^(2))/(r^(2))` <br> `=((8.99xx10^(9)N*m^(2)//C^(2))(1.602xx10^(19)C)^(2))/((4.0xx10^(-5)m)^(2))` <br> `=14N. ` (Answer) <br> No explosion: This is a small force to be acting on a macroscopic object like a cantaloupe, but an enormous force to be acting on a proton. Such forces should explode the nucleus of any element but hydrogen (which has only one proton in its nucleus). However, they don.t not even in nuclei witha great many protons. Therefore, there must be some enormous attractive force to counter this enormous repulsive electrostatic force.

Answer

Verified

If both the charges are positive or both the charges are negative, then the two charges will repel each other. We know that protons are positively charged. Therefore, the two will repel each other. The magnitude of the electrostatic force between two charges ${{q}_{1}}$, ${{q}_{2}}$, which are separated by a distance r is given as $F=\dfrac{K{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ ….. (i), where K is a proportionality constant whose value is $9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$ in free space.The charge on the proton is $1.6\times {{10}^{-19}}C$. Therefore, ${{q}_{1}}={{q}_{2}}=1.6\times {{10}^{-19}}C$It is given that the protons are separated by a distance of 20cm. Therefore, $r=20cm=0.2m$.Substitute the values in (i). $\Rightarrow F=\dfrac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times 1.6\times {{10}^{-19}}}{{{0.2}^{2}}}=5.76\times {{10}^{-27}}N.$

Note:

If the charges were of opposite nature, i.e. one positive and one negative then the two charges will attract each other. Consider the two charges to be a proton and an electron separated by the same distance. The magnitude of charge on an electron is equal to the charge on a proton. Therefore, the two charges will attract each other with the same magnitude of force that was in the case of two protons.

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