Question

Two stationary point charges +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 two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00 nC charge?

Short answer

The speed of an electron when it is 10 cm away from q₁ is 6.89 x 10⁶ m/s.

Step-by-step solution

Given Data and concept

Mass of electron, m_e = 9.11 × 10^-31 kg.

Charge on electron, q_e = -1.60 ×10^-19 C.

Stationary charge 1, q₁ = 3.00 × 10^-9 C.

Stational charge 2, q₂ = 2.00 × 10^-9 C.

r_i = 0.250 m

r_1f = 0.100 m

r_2f = 0.400 m

The Law of energy conservation tells us that the total energy of a system is always conserved . It can not be gained or lost. It can change forms.

Determination of the speed of the electron

Initially, the charges are at the stationary position, so net initial energy is given by,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{i}}={\mathrm{U}}_{\mathrm{i}}+{\mathrm{K}}_{\mathrm{i}} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{{\mathrm{q}}_{\mathrm{e}}{\mathrm{q}}_1}{{\mathrm{r}}_{\mathrm{i}}}+\frac{{\mathrm{q}}_{\mathrm{e}}{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}}\right)+0 \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\left(-1.6\times {10}^{-19}\right)\left(3\times {10}^{-9}\right)}{0.250}+\frac{\left(-1.6\times {10}^{-19}\right)\left(2\times {10}^{-9}\right)}{0250}\right) \\ =-2.88\times {10}^{-17}\mathrm{J} \end{gathered}$$

Finally, the charge starts moving with a velocity, so net final energy is given by,

$$\begin{gathered} E_f=U_f+K_f \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q_e{q}_1}{r_{1f}}+\frac{q_e{q}_2}{r_{2f}}\right)+\frac{1}{2}{m}_e{v}_f^2 \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\left(-1.6\times {10}^{-19}\right)\left(3\times {10}^{-9}\right)}{0.100}+\frac{\left(-1.6\times {10}^{-19}\right)\left(2\times {10}^{-9}\right)}{0.400}\right)+\frac{1}{2}{m}_e{v}_f^2 \\ =-5.04\times {10}^{-17}J+\frac{1}{2}{m}_e{v}_f^2 \end{gathered}$$

Here, v_f is the velocity of the electron.

By Law of conservation of energy,

$$\begin{gathered} E_i=E_f \\ {U}_i+K_i=U_f+k_f \\ -2.80\times {10}^{-17}=-5.04\times {10}^{-17}+\frac{1}{2}{m}_e{v}_f^2 \\ {v}_f=\sqrt{\frac{2\left(-2.88\times {10}^{-17}+5.04\times {10}^{-17}\right)}{9.11\times {10}^{-31}}} \\ {v}_f=6.89\times {10}^{6}m/s \end{gathered}$$

Therefore, the speed of an electron when it is 10 cm away from q₁ is 6.8910⁶ m/s.