Question

Calculate the distance between two protons such that the electrical

repulsive force between them is equal to the weight of either proton.

Given charge on each proton=1.6×10-19C, mass of proton =1.6× 10−27kg,

acceleration due to gravity g= 10m/s2​

Answer 1

mass of proton , m= 1.6726 × 10^-27 Kg

so, weight of each proton , F = mg , where g is acceleration due to gravity.

F = 1.6726 × 10^-27 × 10 = 1.6726 × 10^-26 N

we also know, charge on proton , e = 1.6 × 10^-19 C

so, electrostatic force between protons , F = Ke²/ r² [ by Coulomb's law ]

where K is kappa constant , 9 × 10^9 Nm²/C² and r is the separation between them.

now, weight of proton = electrostatic force

Ke²/r² = 1.6726 × 10^-26

or, 9 × 10^9 × (1.6 × 10^-19)²/r² = 1.6726 × 10^-26

or, r² = 9 × 2.56 × 10^-29/1.6726 × 10^-26

or, r = 0.12 m

hence, distance between two protons = 0.12m

Answer 2

The distance between two protons would be 0.12 m.

Given:

The Repulsive force between two protons = The weight of either proton

i.e., $F_{R}$ = mg

The charge on each proton (q) = 1.6 × $10^{-19}$ C.

The mass of each proton (m) = 1.6 × $10^{-27}$ Kg.

The acceleration due to gravity (g) = 10 $\frac{m}{s^{2} }$.

To Find:

The distance (d) between two protons such that the electrical repulsive force between them is equal to the weight of either proton =?

Solution:

The Electrical force between two charges each of charge Q and q, r distance apart from each other is given by;

⇒ F = k $\frac{Qq}{r^{2} }$ ; where k = 9 × $10^{9}$ $\frac{Nm^{2} }{C^{2} }$

Also, this force is attractive in nature for the opposite nature of charges and is repulsive in nature when charges are of the same nature.

Now, according to the given question,

The repulsive force between two protons is equal to the weight of the proton;

∴  mg  =  k$\frac{q^{2} }{r^{2} }$

∴  (1.6 × $10^{-27}$)(10) = (9 × $10^{9}$ )( 1.6 × $10^{-19}$ )( 1.6 × $10^{-19}$  ) / $r^{2}$

∴  $r^{2}$ = (9 × $10^{9}$ )( 1.6 × $10^{-19}$ )( 1.6 × $10^{-19}$  ) / (1.6 × $10^{-27}$)(10)

∴  $r^{2}$ = 9 × 1.6 × $10^{-3}$

∴  $r^{2}$ = 9 × 16 × $10^{-4}$

∴  $r^{2}$ = 144  × $10^{-4}$

∴ r = 12 × $10^{-2}$

∴ r = 0.12 m.

Therefore, the distance between two protons is equal to 0.12 m.

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