Question

An electron is found to repel another electron at a distance of \(1\,\text{cm}\) with a force \(F=2.3\times10^{-24}\,\text{N}\). Two protons placed at a distance of \(5\,\text{cm}\) will

• A. repel each other with a force \(F=2.3\times10^{-24}\,\text{N}\)
• B. attract each other with a force \(F=2.3\times10^{-24}\,\text{N}\)
• C. repel each other with a force \(F=9.2\times10^{-26}\,\text{N}\)
• D. attract each other with a force \(F=4.6\times10^{-25}\,\text{N}\)

Hint:

For Coulomb force: \[ F\propto \frac{1}{r^2} \] If distance becomes \(n\) times, \[ F \rightarrow \frac{F}{n^2} \] Also, \[ \text{Like charges repel, unlike charges attract.} \]

Answer 1

The Correct Option is C

Concept: According to Coulomb's law, \[ F=\frac{kq_1q_2}{r^2} \] For electrons and protons, the magnitude of charge is the same: \[ |q_e|=|q_p|=e \] Hence the force varies only with the inverse square of distance.

Step 1: Write the given force. For two electrons separated by \[ r_1=1\,\text{cm} \] the force is \[ F_1=2.3\times10^{-24}\,\text{N} \]

Step 2: Find the force at \(5\,\text{cm}\). For two protons, \[ r_2=5\,\text{cm} \] Using \[ \frac{F_2}{F_1} = \frac{r_1^2}{r_2^2} \] \[ F_2 = F_1\left(\frac{1}{5}\right)^2 \] \[ F_2 = \frac{2.3\times10^{-24}}{25} \] \[ F_2 = 9.2\times10^{-26}\,\text{N} \]

Step 3: Determine the nature of force. Since both particles are protons, \[ (+e)\ \text{and}\ (+e) \] the force is repulsive. \[ \boxed{\text{Force is repulsive.}} \]

Step 4: State the answer. \[ \boxed{ F=9.2\times10^{-26}\,\text{N} } \] and the force is repulsive. Hence, the correct option is \[ \boxed{(C)} \]