In Fig. 22-69, particle 1 of charge${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.00}\mathbf{\text{pC}}$and particle 2 of charge${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.00}\mathbf{\text{pC}}$are fixed at a distance$\mathit{d}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.00}\mathbf{}\mathit{c}\mathit{m}$apart. In unit-vector notation, what is the net electric field at points

(a) A,

(b) B, and

(c) C?

(d) Sketch the electric field lines.

In Fig., particle 1 of charge $q_1=1.00\text{\hspace{0.17em}pC}$and particle 2 of charge $q_2=2.00\text{\hspace{0.17em}pC}$are fixed at a distance $d=5.0\text{\hspace{0.17em}cm}$apart.

Using the basic concept of unit-vector notation of a position vector in the concept of the electric field, we get the net electric field acting on a point due to the contribution of present charges.

Formula:

The electric field on a particle due to charge, $\overrightarrow{E}=\frac{q}{4\pi {\epsilon }_o{r}^2}\widehat{r}$ (i)

where, r = the distance of field point from the charge

q = charge of the particle

For point A, using the given data in equation (i), we get the electric field value as given:

$\begin{array}{c}\overrightarrow{E}{}_A=[\frac{q_1}{4\pi {\epsilon }_o{r}_1^2}+\frac{q_2}{4\pi {\epsilon }_o{r}_2^2}](-\widehat{i})\\ =\frac{(8.99\times {10}^9\text{N}\cdot \frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00\times {10}^{-12}C)}{{(5.00\times {10}^{-2}\text{\hspace{0.17em}C})}^2}(-\widehat{i})+\frac{(8.99\times {10}^9\text{N}\cdot \frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00\times {10}^{-12}\text{\hspace{0.17em}C})}{{(5.00\times {10}^{-2}\text{\hspace{0.17em}C})}^2}(+\widehat{i})\\ =(-1.80\text{N/C})\widehat{i}\end{array}$

Hence, the value of the force is.$(-1.80\text{N/C})\widehat{i}$

For point B, using the given data in equation (i), we get the electric field value as given:

.$\begin{array}{c}\overrightarrow{E}{}_B=[\frac{q_1}{4\pi {\epsilon }_o{r}_1^2}+\frac{\left|q_2\right|}{4\pi {\epsilon }_o{r}_2^2}]\left(\widehat{i}\right)\\ =\frac{(8.99\times {10}^9\text{N}\cdot \frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00\times {10}^{-12}\text{\hspace{0.17em}C})}{{(5.00\times {10}^{-2}\text{\hspace{0.17em}C})}^2}\left(\widehat{i}\right)+\frac{(8.99\times {10}^9\text{N}\cdot \frac{\text{m}}{{\text{C}}^{\text{2}}})|-2.00\times {10}^{-12}\text{\hspace{0.17em}C}|}{{(5.00\times {10}^{-2}\text{\hspace{0.17em}C})}^2}(+\widehat{i})\\ =\left(43.2\text{N/C}\right)\widehat{i}\end{array}$

Hence, the value of the force is.$\left(43.2\text{N/C}\right)\widehat{i}$

For point C, using the given data in equation (i), we get the electric field value as given:

$\begin{array}{c}\overrightarrow{E}{}_C=[\frac{q_1}{4\pi {\epsilon }_o{r}_1^2}-\frac{\left|q_2\right|}{4\pi {\epsilon }_o{r}_2^2}]\left(\widehat{i}\right)\\ =\frac{(8.99\times {10}^9\text{N}\cdot \frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00\times {10}^{-12}\text{\hspace{0.17em}C})}{{(5.00\times {10}^{-2}\text{\hspace{0.17em}C})}^2}\left(\widehat{i}\right)-\frac{(8.99\times {10}^9\text{N}\cdot \frac{\text{m}}{{\text{C}}^{\text{2}}})|-2.00\times {10}^{-12}\text{\hspace{0.17em}C}|}{{(5.00\times {10}^{-2}\text{\hspace{0.17em}C})}^2}(+\widehat{i})\\ =-\left(6.29\text{N/C}\right)\widehat{i}\end{array}$

Hence, the value of the force is$-\left(6.29\text{N/C}\right)\widehat{i}$.

The field lines are shown to the right. Note that there are twice as many field lines “going into” the negative charge$-2q$as compared to that flowing out from the positive charge.$+q$