Q4P - Fundamentals Of Physics

Question

Two charged particles are attached to an x-axis: Particle 1 of charge $\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{C}$ is at position $\mathbf{x}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{cm}$ and particle 2 of charge +  $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{C}$ is at position. Midway between the particles, what is their net electric field in unit-vector notation?

Step-by-Step Solution

Answer

The net electric field in unit-vector notation midway between the particles is $(-6.39\times {10}^5 \mathrm{N}/\mathrm{C})\hat{\mathrm{i}}$

Step 1: The given data

(a)   Charge of particle 1 at position ${\mathrm{x}}_1=6\hspace{0.33em}\mathrm{cm}\left(\frac{1\hspace{0.33em}\mathrm{m}}{100 \mathrm{cm}}\right)=0.06\hspace{0.33em}\mathrm{m}$ is ${\mathrm{q}}_1=–2.00\mathrm{x}{10}^{-7}\mathrm{C}.$

(b)   Charge of particle 2 at position ${\mathrm{x}}_2=21\hspace{0.33em}\mathrm{cm}\left(\frac{1\hspace{0.33em}\mathrm{m}}{100 \mathrm{cm}}\right)=0.21\hspace{0.33em}\mathrm{m}$ is ${\mathrm{q}}_2=+2.00\mathrm{x}{10}^{-7}\hspace{0.33em}\mathrm{C}.$

(c)   The midway point between the charges is located at $\mathrm{x}=13.5 \mathrm{cm}\left(\frac{ 1\hspace{0.33em}\mathrm{m}}{100 \mathrm{cm}}\right)= 0.135\hspace{0.33em}\mathrm{m}$

Step 2: Understanding the concept of electric field

Using the concept of the electric field, the value of the net midway electric field can be calculated by using the unit vector notation.

 

Formula:

The formula of the electric field in unit-vector notation,$\overrightarrow{\mathrm{E}}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}\hat{\mathrm{R}}$      (1)

where R = The distance of field point from the charge and q = charge of the particle

Step 3: Calculation of the net electric field in unit-vector notation

The magnitudes and directions of the individual fields for particles 1 and 2 using equation (i) are given as follows:

$$\begin{gathered} \overrightarrow{{\mathrm{E}}_1}=-\frac{\left(8.99\times {10}^9 \mathrm{N}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{C}}^2}}\right)\left(2.00\times {10}^{-7 }{\displaystyle \frac{\mathrm{N}}{\mathrm{C}}}\right)}{{\left(0.135 \mathrm{m}-0.060 \mathrm{m}\right)}^2}\hat{\mathrm{i}} \\ =\left(-3.196\times {10}^5 \mathrm{N}/\mathrm{C}\right)\hat{\mathrm{i}} \end{gathered}$$

.

  $$\begin{gathered} \overrightarrow{{\mathrm{E}}_2}=-\frac{\left(8.99\times {10}^9 \mathrm{N}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{C}}^2}}\right)\left(2.00\times {10}^{-7 }{\displaystyle \frac{\mathrm{N}}{\mathrm{C}}}\right)}{{\left(0.135 \mathrm{m}-0.210 \mathrm{m}\right)}^2}\hat{\mathrm{i}} \\ =-\left(3.196\times {10}^5 \mathrm{N}/\mathrm{C}\right)\hat{\mathrm{i}} \end{gathered}$$

Thus, using the above two electric field values, the net field of the particles is given as:

$$\begin{gathered} \overrightarrow{{\mathrm{E}}_{\mathrm{net}}}=\overrightarrow{{\mathrm{E}}_1}+\overrightarrow{{\mathrm{E}}_2} \\ -(6.39\times {10}^5 \mathrm{N}/\mathrm{C})\hat{\mathrm{i}} \end{gathered}$$

Hence, the value of the electric field is $-(6.39\times {10}^5 \mathrm{N}/\mathrm{C})\hat{\mathrm{i}}$