Let's carefully analyze the problem: **Problem Restatement:** - Two electrons are initially infinitely far apart, so their total mass is just \( 2m_e \) (where \( m_e \) is the electron mass). - When they are brought closer together, their total *relativistic mass* (or equivalently, their total energy divided by \( c^2 \)) increases due to their mutual Coulomb potential energy. - The question asks: **How close must they be positioned so that their combined relativistic mass is 5 times their mass at infinite separation?** --- ### Step 1: Understand the concept of mass increase In special relativity, the *relativistic mass* of a system includes the rest mass plus the energy stored in the system. When particles are bound or interacting via potential energy, their total energy (including potential energy) contributes to the system's relativistic mass. **Key points:** - Rest mass of two electrons: \( 2m_e \) - Potential energy of two electrons separated by distance \( r \): \[ U(r) = \frac{k e^2}{r} \] where: - \( e \) is the elementary charge, - \( k = \frac{1}{4\pi\varepsilon_0} \) The total energy of the system at separation \( r \): \[ E(r) = 2 m_e c^2 + U(r) \] The **effective relativistic mass** (or total energy divided by \( c^2 \)): \[ M_{eff}(r) = \frac{E(r)}{c^2} = 2 m_e + \frac{U(r)}{c^2} \] --- ### Step 2: Set the condition for the total mass The problem states that when the electrons are at distance \( r \), their total relativistic mass is **5 times** their mass at infinite separation: \[ M_{eff}(r) = 5 \times 2 m_e = 10 m_e \] Using the expression above: \[ 2 m_e + \frac{U(r)}{c^2} = 10 m_e \] Solve for \( U(r) \): \[ \frac{U(r)}{c^2} = 8 m_e \] --- ### Step 3: Express the Coulomb potential energy Recall: \[ U(r) = \frac{k e^2}{r} \] Substitute into the previous equation: \[ \frac{k e^2}{r c^2} = 8 m_e \] Solve for \( r \): \[ r = \frac{k e^2}{8 m_e c^2} \] --- ### Step 4: Plug in known constants - \( k = \frac{1}{4 \pi \varepsilon_0} \approx 8.987551787 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) - \( e = 1.602176634 \times 10^{-19} \, \text{C} \) - \( m_e = 9.10938356 \times 10^{-31} \, \text{kg} \) - \( c = 2.998 \times 10^{8} \, \text{m/s} \) Calculate numerator first: \[ k e^2 = (8.987551787 \times 10^9) \times (1.602176634 \times 10^{-19})^2 \] Calculate \( e^2 \): \[ (1.602176634 \times 10^{-19})^2 \approx 2.56697 \times 10^{-38} \] Multiply: \[ k e^2 \approx 8.987551787 \times 10^9 \times 2.56697 \times 10^{-38} \approx 2.306 \times 10^{-28} \, \text{J}\cdot \text{m} \] Now, denominator: \[ 8 m_e c^2 = 8 \times 9.10938356 \times 10^{-31} \times (2.998 \times 10^{8})^2 \] Calculate \( c^2 \): \[ (2.998 \times 10^{8})^2 \approx 8.988 \times 10^{16} \] Multiply: \[ 8 \times 9.10938356 \times 10^{-31} \times 8.988 \times 10^{16} \approx 8 \times 9.10938356 \times 8.988 \times 10^{-31 + 16} = 8 \times 9.10938356 \times 8.988 \times 10^{-15} \] Calculate the numerical part: \[ 8 \times 9.10938356 \times 8.988 \approx 8 \times 81.89 \approx 655.12 \] Thus, the denominator: \[ \approx 655.12 \times 10^{-15} = 6.5512 \times 10^{-13} \] --- ### Step 5: Final calculation of \( r \) \[ r = \frac{2.306 \times 10^{-28}}{6.5512 \times 10^{-13}} \approx 3.52 \times 10^{-16} \, \text{m} \] --- ### **Final Answer:** \[ \boxed{ \text{The electrons must be separated by approximately } \quad r \approx 3.5 \times 10^{-16} \text{ meters} } \] This is **about 0.35 femtometers**, which is roughly comparable to the size of a proton, indicating that to increase their total mass to five times the initial, they must be brought extremely close together—on the order of nuclear dimensions.