Question

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.

Short answer

1. The net electric field at point A is $(-1.80\text{N/C})\widehat{i}$
2. The net electric field at point B is $\left(43.2\text{N/C}\right)\widehat{i}$
3. The net electric field at point C is$-\left(6.29\text{N/C}\right)\widehat{i}$
4. The electric field lines are sketched for all the forces at the given points.

Step-by-step solution

The given data

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.

Understanding the concept of the electric field

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

a) Calculation of the electric field at point A

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.$(-1.80\text{N/C})\widehat{i}$

b) Calculation of the electric field at point B

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.$\left(43.2\text{N/C}\right)\widehat{i}$

c) Calculation of the electric field at point C

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$-\left(6.29\text{N/C}\right)\widehat{i}$.

d) Calculation for sketching the electric field lines of the acting forces

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.$+q$