Question

Two protons are separated by a distance of 2.0×10^-15m. What is the magnitude of electric force between them?

Answer 1

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✺ Answer:

♦️ GiveN:

• Charge on 1 proton = 1.6 × 10^-19 C
• Distance between them = 2 ×10^-15 m

♦️ To FinD:

• Magnitude of electrostatic force between the protons.

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✺ Concept Used:

To solve this question, we need to know about the coulombs law, and the formula based upon that.

$\large{ \star{ \underline{ \boxed{ \sf{ \red{Coulombs \: law}}}}}}$

According to this law, the electric static force/ electric force between any two particles is directly proportional to the product of magnitude of charges of these particles and inversely proportional to the square of distance between them.

So, we can write it as,

$\large{ \star{ \boxed{ \rm{F = k \frac{q1q2}{ {r}^{2} }}}}} \\ \normalsize{ \rm{where \: symbols \: have \: their \: usual \: meanings...}}$

Here, k is the constant of proportionality, and,

$\large{ \rm{k = \frac{1}{4\pi e_0 } \: and \: k = 9 \times {10}^{9} N{m}^{2} {C}^{ - 2} }}$

By using this formula, let's find the distance between the two protons....

♠️Refer to the attachment....

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✺ Solution:

As we have all the required values, we can now simply solve by using the formula.

By using formula,

$\large{ \rm{ \dashrightarrow \: F = \frac{9 \times {10}^{9} \times (1.6 \times {10}^{ - 19}) {}^{2} }{(2 \times {10}^{ - 15}) {}^{2} }N}} \\ \\ \large{ \rm{ \dashrightarrow \: F = \frac{9 \times {10}^{9} \times 1.6 \times {10}^{ - 19} \times 1.6 \times {10}^{ - 19} }{2 \times {10}^{ - 15} \times 2 \times {10}^{ - 15} }\:N} } \\ \\ \large{ \rm{ \dashrightarrow \: F = \frac{9 \times 1.6 \times 1.6}{2 \times 2} \times 10} \: N} \\ \\ \large{ \rm{ \dashrightarrow \: F= 5.76 \times 10} \: N} \\ \\ \large{ \rm{ \dashrightarrow \: F = \boxed{ \red{ \rm{57.6 \: N}}}}} \\ \\ \large{ \therefore{ \underline{ \rm{ \green{The \: force \: between \: these \: 2 \: protons = 57.6 \: N}}}}}$

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