Question

Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.

Answer 1

Answer:

We know, Charge on one Proton = + 1.6 × 10⁻¹⁹ C.and Mass of the Proton = 1.6727 × 10⁻²⁷ kg.

Now, Using the Coulomb's Law,

F = (k q₁ q₂) ÷ r²

where,

k = 8.99 × 10⁹ Nm²kg⁻²

k ≈ 9 × 10⁹ Nm²kg⁻²

q₁ = q₂ = 1.6 × 10⁻¹⁹ C

r = Distance between the Two Protons.

Now,

F = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² ÷ r²

⇒ F₁ = 23.04 × 10⁻²⁹/r² -----eq(i)

Now, Using Newton's law of Gravitation,

F = G \frac{m_{1} m_{2} }{r^{2} }F=G

Here, G = Gravitation Constant whose value is 6.67 × 10⁻¹¹ Nm²kg⁻².

⇒ F r² = 6.67 × 10⁻¹¹ × (1.6727 × 10⁻²⁷)²

⇒ F₂ = 18.66 × 10⁻⁶⁵/r²

Now, Let us calculate the ratio of the Electrostatic Forces to the Gravitational Force between the Two Protons.

∴ F₁/F₂ = (23.04 × 10⁻²⁹/r²) ÷ (18.66 × 10⁻⁶⁵/r²)

⇒ F₁/F₂ = 1.24 × 10³⁶

Hence, the Ratio is 1.24 × 10³⁶.

Answer 2

To Find: The ratio of the magnitude of the electric force to the gravitational force acting between two protons.

Explanation:

Step 1:

We know,  

Charge on one Proton =  + 1.6 × 10⁻¹⁹ C.

and  

Mass of the Proton = 1.6727 × 10⁻²⁷ kg.

Step 2:

Now, Using the Coulomb's Law,

$\begin{equation}F=\frac{\left(\mathrm{k} q_{1} \mathrm{q}_{2}\right)}{r^{2}}$

where,

k = 8.99 × 10⁹ Nm²kg⁻²

k ≈ 9 × 10⁹ Nm²kg⁻²

q₁ = q₂ = 1.6 × 10⁻¹⁹ C

r= The Two Protons Range.

Now,

$\begin{equation}F_{1}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{r^{2}}$

$\begin{equation}F_{1}=\frac{23.04 \times 10^{-2}}{r^{2}}$            ......eqn 1

Step 3:

Now, using the law of Gravitation of Newton,

here,

G = Gravity Constant, whose value is 6.67 x $10^{-11}$ $Nm^2$$kg^{-2}$.

$\begin{equation}\begin{array}{c}{F_{2}=\frac{6.67 \times 10^{-11} \times\left(1.6727 \times 10^{-27}\right)^{2}}{r^{2}}} \\{F_{2}=\frac{18.66 \times 10^{-65}}{r^{2}}}\end{array}$

F_2=(18.66 × 10⁻⁶⁵   )/r^2  

Step 4:

Now, let's calculate the ratio between the Two Protons of the Electrostatic Forces to the Gravitational Force.

$\begin{equation}\begin{aligned}\frac{\mathrm{F}_{1}}{\mathrm{F}_{2}} &=\frac{\frac{23.04 \times 10^{-29}}{r^{2}}}{\frac{18.66 \times 10^{-65}}{r^{2}}} \\\frac{\mathrm{F}_{1}}{\mathrm{F}_{2}} &=\frac{23.04 \times 10^{-29}}{18.66 \times 10^{-65}} \\\frac{\mathrm{F}_{1}}{\mathrm{F}_{2}} &=1.24 \times 10^{36}\end{aligned}$