**Speed of an Electron in an Electric Field**

**Solution Below:**

Part A

Express your answer in meters per second.

This problem is kind of complicated. You can solve it by finding the difference between the potential energy of the electron at the beginning and the end and converting that to kinetic energy (and from there to velocity). Start by finding the potential energy of the electron midway between the two charges (the radius will be 0.57m / 2 = 0.285m):

V_{i} = kQ_{1}/r + kQ_{2}/r

V_{i} = k(Q_{1} + Q_{2})/r

V_{i} = k * (4.00nC + 1.95nc)/ 0.285

Convert to coulombs (from nC) and solve:

V_{i} = 2.08772 * 10^-8 * k

V_{i} = 2.08772 * 10^-8 * (8.998 * 10^9)

V_{i} = 187.853 V

Now find the beginning potential energy of the electron:

U_{i} = q * V_{i}

U_{i} = 1.602*10^-19 * 187.853

U_{i} = 3.0094*10^-17

Now do the same thing, but for the position of the electron that this problem is asking for (10.0 cm from charge 1). The radius will be different for each charge this time around:

V_{f} = kQ_{1}/r_{1} + kQ_{2}/r_{2}

V_{f} = k * 4.00nC / 0.1 + k * 1.95nC / 0.47

Convert to coulombs again and solve:

V_{f} = 397.252 V

Now find the final potential energy of the electron:

U_{f} = q * V_{f}

U_{f} = 1.602*10^-19 * 397.252

U_{f} = 6.3640 * 10^-17

Now solve for kinetic energy (use the absolute value of the difference in potential energies):

KE = abs(U_{i} – U_{f})

KE = abs((3.0094*10^-17) – (6.3640 * 10^-17))

KE = 3.3546 * 10^-17

Convert to velocity (the mass of an electron is 9.1094 * 10^-31 kg):

KE = 1/2mv^2

3.3546 * 10^-17 = 1/2mv^2

6.7092 * 10^-17 = mv^2

v^2 = 7.3651 * 10^13

v = 8,577,000 m/s (difference due to rounding)

8,577,000 m/s