Question

The electric field at a point in space is$\overrightarrow{E}=(400\hat{i}+100\hat{j})\mathrm{N}/\mathrm{C}$

a. What is the electric force on a proton at this point? Give your answer in component form.

b. What is the electric force on an electron at this point? Give your answer in component form.

c. What is the magnitude of the proton’s acceleration?

d. What is the magnitude of the electron’s acceleration?

Short answer

a. force on a proton $=(6.4\hat{i}+1.6\hat{j})\times {10}^{-17}\mathbf{N}$

b. force on a electron $=-(6.4\hat{i}+1.6\hat{j})\times {10}^{-17}\mathbf{N}$

c. Proton's acceleration $=4\times {10}^{10}m/s^2$

d. Electron's acceleration$=7.3\times {10}^{13}m/s^2$

Step-by-step solution

Given information

The electric field at a point in space is $\overrightarrow{E}=(400\hat{i}+100\hat{j})\mathrm{N}/\mathrm{C}$

If a proton or electron is placed in space, the magnitude of the proton's acceleration and electron's acceleration is needed.

$protonmass=1.67262192\times {10}^{-27}kg$

$electronmass=9.10938\times {10}^{-31}kg$

The charge of the electron: $e=-1.6\times {10}^{-19}\mathrm{C}$

The charge of the proton:$p=1.6\times {10}^{-19}C$

Explanation (part a)

The magnitude of the electric force (F) experienced by the proton is computed as:

$\begin{array}{r}F=\left|Ep\right|\\ =\left|(400\hat{i}+100\hat{j})\frac{\mathrm{N}}{\mathrm{C}}\times \left(1.6\times {10}^{-19}\mathrm{C}\right)\right|\\ =(6.4\hat{i}+1.6\hat{j})\times {10}^{-17}\mathbf{N}\end{array}$

Explanation (part b)

The magnitude of the electric force (F) experienced by the electron is computed as:

$\begin{array}{r}F=Ee\\ =(400\hat{i}+100\hat{j})\frac{\mathrm{N}}{\mathrm{C}}\times \left(-1.6\times {10}^{-19}\mathrm{C}\right)\\ =-(6.4\hat{i}+1.6\hat{j})\times {10}^{-17}\mathbf{N}\end{array}$

Explanation (part c)

The magnitude of the acceleration of the proton at that point is calculated as: $a_p=\frac{F_{net}}{m_p}$

The net force of the proton is calculated as $F_{net}=\sqrt{({(6.4\times {10}^{-17})}^2+{(1.6\times {10}^{-17})}^2)}=6.59\times {10}^{-17}N$

Plugging the above values in the above equation, we get proton's acceleration as:

$a_p=\frac{6.59\times {10}^{-17}}{1.7\times {10}^{-27}}=4\times {10}^{10}m/s^2$

Explanation (part d)

The magnitude of the acceleration of the electron at that point is calculated as: $a_e=\frac{F_{net}}{m_e}$

The net force of the electron is calculated as

$F_{net}=\sqrt{({(-6.4\times {10}^{-17})}^2+{(-1.6\times {10}^{-17})}^2)}=6.59\times {10}^{-17}N$

Plugging the above values in the above equation, we get electron's acceleration as:

$a_e=\frac{6.59\times {10}^{-17}}{9.1\times {10}^{-31}}=7.3\times {10}^{13}m/s^2$