91影视

In Fig., particle 1 of charge $q_1=1.00\text{鈥塸颁}$and particle 2 of charge $q_2=2.00\text{鈥塸颁}$are fixed at a distance $d=5.0\text{鈥塩尘}$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, $\stackrel{鈫�}{E}=\frac{q}{4蟺{蔚}_o{r}^2}\widehat{r}$ (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:

$\begin{array}{c}\stackrel{鈫�}{E}{}_A=[\frac{q_1}{4蟺{蔚}_o{r}_1^2}+\frac{q_2}{4蟺{蔚}_o{r}_2^2}](鈭�\widehat{i})\\ =\frac{(8.99脳{10}^9\text{N}鈰�\frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00脳{10}^{鈭�12}C)}{{(5.00脳{10}^{鈭�2}\text{鈥塁})}^2}(鈭�\widehat{i})+\frac{(8.99脳{10}^9\text{N}鈰�\frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00脳{10}^{鈭�12}\text{鈥塁})}{{(5.00脳{10}^{鈭�2}\text{鈥塁})}^2}(+\widehat{i})\\ =(鈭�1.80\text{N/C})\widehat{i}\end{array}$

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

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

.$\begin{array}{c}\stackrel{鈫�}{E}{}_B=[\frac{q_1}{4蟺{蔚}_o{r}_1^2}+\frac{\left|q_2\right|}{4蟺{蔚}_o{r}_2^2}]\left(\widehat{i}\right)\\ =\frac{(8.99脳{10}^9\text{N}鈰�\frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00脳{10}^{鈭�12}\text{鈥塁})}{{(5.00脳{10}^{鈭�2}\text{鈥塁})}^2}\left(\widehat{i}\right)+\frac{(8.99脳{10}^9\text{N}鈰�\frac{\text{m}}{{\text{C}}^{\text{2}}})|鈭�2.00脳{10}^{鈭�12}\text{鈥塁}|}{{(5.00脳{10}^{鈭�2}\text{鈥嬧赌塁})}^2}(+\widehat{i})\\ =\left(43.2\text{N/C}\right)\widehat{i}\end{array}$

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:

$\begin{array}{c}\stackrel{鈫�}{E}{}_C=[\frac{q_1}{4蟺{蔚}_o{r}_1^2}鈭�\frac{\left|q_2\right|}{4蟺{蔚}_o{r}_2^2}]\left(\widehat{i}\right)\\ =\frac{(8.99脳{10}^9\text{N}鈰�\frac{\text{m}}{{\text{C}}^{\text{2}}})(1.00脳{10}^{鈭�12}\text{鈥塁})}{{(5.00脳{10}^{鈭�2}\text{鈥塁})}^2}\left(\widehat{i}\right)鈭�\frac{(8.99脳{10}^9\text{N}鈰�\frac{\text{m}}{{\text{C}}^{\text{2}}})|鈭�2.00脳{10}^{鈭�12}\text{鈥塁}|}{{(5.00脳{10}^{鈭�2}\text{鈥塁})}^2}(+\widehat{i})\\ =鈭�\left(6.29\text{N/C}\right)\widehat{i}\end{array}$

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

The field lines are shown to the right. Note that there are twice as many field lines 鈥済oing into鈥� the negative charge$鈭�2q$as compared to that flowing out from the positive charge.$+q$