Answer to Question #126625 in Physics for p

Question

Two particles are in a uniform electric field whose value is +2500 N/C. The mass and charge of particle 1 are m1 = 1.08 × 10-5 kg and q1 = -7.47 uC, while the corresponding values for particle 2 are m2 = 2.67 × 10-5 kg and q2 = +18.4 uC. Initially the particles are at rest. The particles are both located on the same electric field line but are separated from each other by a distance d. When released, they accelerate, but always remain at this same distance from each other. Find d.

Answer 1

The acceleration of each charge must be the same if the distance between the charges remains unchanged:

$$a_1=a_2\\\frac{q_1E-k\frac{q_1q_2}{d^2}}{m_1}=\frac{q_2E+k\frac{q_1q_2}{d^2}}{m_2}\\\frac{(7.47\cdot10^{-6})(2500)-(8.99\cdot10^{9})\frac{(7.47\cdot10^{-6})(18.4\cdot10^{-6})}{d^2}}{1.08\cdot10^{-5}}\\=\frac{(18.4\cdot10^{-6})(2500)+(8.99\cdot10^{9})\frac{(7.47\cdot10^{-6})(18.4\cdot10^{-6})}{d^2}}{2.67\cdot10^{-5}}$$

$$d=159\ m$$