Question

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy of the electron when it is
1. at the midpoint?
2. 10.0 cm from the 3.00 nC charge?

Answer 1

Answer:

1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  $10^{-17}$ J

Explanation:

given information:

$q_{1}$ =  3 nC = 3 x $10^{-9}$ C

$q_{2}$ =  2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$)

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

  = 0.25 m

thus,

F = $F_{1}+ F_{2}$

  = k$\frac{q_{e} q_{1} }{r}$ + k$\frac{q_{e} q_{2} }{r}$

  = $\frac{kq_{e} }{r}$ ($q_{1} +q_{2}$)

  = (8.99 x $10^{9}$)( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$)/0.25

  = - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k$\frac{q_{e} q_{1} }{r}$ + k$\frac{q_{e} q_{2} }{r}$

  = $kq_{e}$($\frac{q_{1} }{r_{1} }$+$\frac{q_{2} }{r_{2} }$)

  = (8.99 x $10^{9}$)( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$/0.4)

  = - 5.04 x  $10^{-17}$ J