Question

If the uniform upward-pointing, electric field depicted in Figure 13.44 has a magnitude of $\mathbf{5000}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$, what is the magnitude of the force on the electron while it is in the box? If a different particle experiences a force of$\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}\mathbf{}}\mathbf{N}$ when passing through this region, what is the charge of the particle?

Short answer

The magnitude of the force on the electron while it is in the box is$8\times {10}^{-16}\mathrm{N}$.

The magnitude of the charge on the particle is $3.2\times {10}^{-19}\mathrm{C}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The magnitude of electric field is, $\mathrm{E}=5000\mathrm{N}/\mathrm{C}$.
• The magnitude of electric force on the particle is,${\mathrm{F}}_1=1.6\times {10}^{-15}\mathrm{N}$.

Significance of electric dipole

Whenever a different charged particle lies in a particular electric field, each charge encounters a different value of electric force. The electric force has a direct relationship with the electric charge as well as the electric field.

Determination of the magnitude of the force on the electron while it is in the box.

The expression to calculate the magnitude of the force on the electron while it is on the box can be expressed as,

$\mathrm{F}={\mathrm{q}}_{\mathrm{e}}\mathrm{E}$

Here, F represents the magnitude of the force on the electron while it is on the box and${\mathrm{q}}_{\mathrm{e}}$ represents the electric charge on an electron whose value is $1.6\times {10}^{-19}\mathrm{C}$.

Substitute all the values in the above equation.

$\begin{array}{rcl}\mathrm{F}& =& \left(1.6\times {10}^{-19}\mathrm{C}\right)\left(5000\mathrm{N}/\mathrm{C}\right)\\ & =& 8\times {10}^{-16}\mathrm{N}\end{array}$

Hence, the magnitude of the force on the electron while it is on the box is $8\times {10}^{-6}\mathrm{N}$.

Determination of the charge of the particle.

The expression to calculate the charge on the particle can be expressed as,

$$\begin{gathered} {\mathrm{F}}_1=\mathrm{qE} \\ \mathrm{q}=\frac{{\mathrm{F}}_1}{\mathrm{E}} \end{gathered}$$

Here, q represents the charge on the particle.

Substitute all the values in the above equation.

$\begin{array}{rcl}\mathrm{q}& =& \frac{1.6\times {10}^{-15}\mathrm{N}}{5000\mathrm{N}/\mathrm{C}}\\ & =& 3.2\times {10}^{-19}\mathrm{C}\end{array}$

Hence, the magnitude of the charge on the particle is $3.2\times {10}^{-19}\mathrm{C}$.