91Ó°ÊÓ

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

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

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

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,$\stackrel{→}{\mathrm{E}}=\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}{\mathrm{R}}^2}\hat{\mathrm{R}}$ (1)

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

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

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

.

role="math" localid="1657282764496" $$\begin{gathered} \stackrel{→}{{\mathrm{E}}_2}=-\frac{\left(8.99×{10}^9\mathrm{N}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{C}}^2}}\right)\left(2.00×{10}^{-7}{\displaystyle \frac{\mathrm{N}}{\mathrm{C}}}\right)}{{\left(0.135\mathrm{m}-0.210\mathrm{m}\right)}^2}\hat{\mathrm{i}} \\ =-\left(3.196×{10}^5\mathrm{N}/\mathrm{C}\right)\hat{\mathrm{i}} \end{gathered}$$

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

role="math" localid="1657282406253" $$\begin{gathered} \stackrel{→}{{\mathrm{E}}_{\mathrm{net}}}=\stackrel{→}{{\mathrm{E}}_1}+\stackrel{→}{{\mathrm{E}}_2} \\ -(6.39×{10}^5\mathrm{N}/\mathrm{C})\hat{\mathrm{i}} \end{gathered}$$

Hence, the value of the electric field isrole="math" localid="1657282399133" $-(6.39×{10}^5\mathrm{N}/\mathrm{C})\hat{\mathrm{i}}$