91影视

Amount of charges $=q_1=q_2=q_3=1.0nC=1.0脳{10}^{-9}C$

Distance between the charges is given as $1.0cm=0.01m$

Distance of the point from charge 1 and charge 3 localid="1648553592592" $=r_1=r_3=\sqrt{3.0^2+1.0^2}=3.16cm=0.0316m$

Distance of the point from charge 2$=r_2=3.0cm=0.03m$

Electric field at a point is given by the equation:

$\stackrel{鈫�}{E}=\frac{1}{4蟺{系}_0}\frac{q}{r^2}\hat{r}$

Hence,

for charge 1, the magnitude of the electric field at the given point will be:

$$\begin{gathered} E_1=\frac{1}{4蟺{系}_0}\frac{\left|q\right|}{{r_1}^2} \\ {E}_1=\frac{9.0脳{10}^9脳1.0脳{10}^{-9}}{{\displaystyle 0.{0316}^2}} \\ {E}_1=9.0脳{10}^{3}N/C \end{gathered}$$

In component form, the electric potential is given as:

role="math" localid="1648554529608" ${\stackrel{鈫�}{E}}_1=E_1\cos 胃\hat{i}+E_1\sin 胃\hat{j}$

By the geometry of the given figure, for charge 1,

$$\begin{gathered} \cos 胃=3.0/3.16=0.95 \\ \sin 胃=1.0/3.16=0.32 \end{gathered}$$

Hence, the electric potential can be calculated as:

role="math" localid="1648554573131" $$\begin{gathered} {\stackrel{鈫�}{E}}_1=9.0脳{10}^3脳0.95\hat{i}+9.0脳{10}^3脳0.32\hat{j} \\ {\stackrel{鈫�}{E}}_1=8544.3\hat{i}+2848\hat{j}N/C \end{gathered}$$

Similarly,

for charge 2, the magnitude of the electric field at the given point will be:

role="math" localid="1648554778663" $$\begin{gathered} E_2=\frac{1}{4蟺{系}_0}\frac{\left|q\right|}{{r_2}^2} \\ {E}_2=\frac{9.0脳{10}^9脳1.0脳{10}^{-9}}{{\displaystyle 0.{03}^2}} \\ {E}_2=10.0脳{10}^{3}N/C \end{gathered}$$

In component form, the electric potential is given as:

${\stackrel{鈫�}{E}}_2=E_2\cos 胃\hat{i}+E_2\sin 胃\hat{j}$

By the geometry of the given figure, for charge 2, since there is no angle i.e. the angle is zero,

$$\begin{gathered} \cos 胃=1 \\ \sin 胃=0 \end{gathered}$$

Hence, the electric potential can be calculated as:

$$\begin{gathered} {\stackrel{鈫�}{E}}_2=10.0脳{10}^3\hat{i}+0\hat{j} \\ {\stackrel{鈫�}{E}}_2=10000\hat{i}N/C \end{gathered}$$

Similarly,

for charge 3, the magnitude of the electric field at the given point will be:

$$\begin{gathered} E_3=\frac{1}{4蟺{系}_0}\frac{\left|q\right|}{{r_3}^2} \\ {E}_3=\frac{9.0脳{10}^9脳1.0脳{10}^{-9}}{{\displaystyle 0.{0316}^2}} \\ {E}_3=9.0脳{10}^{3}N/C \end{gathered}$$

In component form, since charge 3 is on the downward side of the point, the electric potential is given as:

role="math" localid="1648555266700" ${\stackrel{鈫�}{E}}_3=E_3\cos 胃\hat{i}-E_3\sin 胃\hat{j}$

By the geometry of the given figure, for charge 3,

$$\begin{gathered} \cos 胃=3.0/3.16=0.95 \\ \sin 胃=1.0/3.16=0.32 \end{gathered}$$

Hence, the electric potential can be calculated as:

$$\begin{gathered} {\stackrel{鈫�}{E}}_3=9.0脳{10}^3脳0.95\hat{i}-9.0脳{10}^3脳0.32\hat{j} \\ {\stackrel{鈫�}{E}}_3=8544.3\hat{i}-2848\hat{j}N/C \end{gathered}$$

Hence, the electric fields created by the three given charges are as follows:

$$\begin{gathered} {\stackrel{鈫�}{E}}_1=8544.3\hat{i}+2848\hat{j}N/C, \\ {\stackrel{鈫�}{E}}_2=10000\hat{i}N/C, \\ {\stackrel{鈫�}{E}}_3=8544.3\hat{i}-2848\hat{j}N/C \end{gathered}$$

Amount of charges$=q_1=q_2=q_3=1.0nC=1.0脳{10}^{-9}C$

Distance between the charges is given as$1.0cm=0.01m$

Distance of the point from charge 1 and charge 3$=r_1=r_3=\sqrt{3.0^2+1.0^2}=3.16cm=0.0316m$

Distance of the point from charge 2$=r_2=3.0cm=0.03m$

The electric fields created by the three given charges are calculated as:

$$\begin{gathered} {\stackrel{鈫�}{E}}_1=8544.3\hat{i}+2848\hat{j}N/C, \\ {\stackrel{鈫�}{E}}_2=10000\hat{i}N/C, \\ {\stackrel{鈫�}{E}}_3=8544.3\hat{i}-2848\hat{j}N/C \end{gathered}$$

From these calculated electric fields, it can be seen that the vertical component i.e. the y-component of the first and the third fields are equal but opposite in nature. Hence they tend to cancel out each other. As a result, there is no net electric field in the vertical direction. Therefore, they obey the principle of superposition.

Thus, the electric fields obey the principle of superposition.

Amount of charges$=q_1=q_2=q_3=1.0nC=1.0脳{10}^{-9}C$

Distance between the charges is given as$1.0cm=0.01m$

Distance of the point from charge 1 and charge 3$=r_1=r_3=\sqrt{3.0^2+1.0^2}=3.16cm=0.0316m$

Distance of the point from charge 2$=r_2=3.0cm=0.03m$

The electric fields created by the three given charges are calculated as:

$$\begin{gathered} {\stackrel{鈫�}{E}}_1=8544.3\hat{i}+2848\hat{j}N/C, \\ {\stackrel{鈫�}{E}}_2=10000\hat{i}N/C, \\ {\stackrel{鈫�}{E}}_3=8544.3\hat{i}-2848\hat{j}N/C \end{gathered}$$

The net electric field is given as:

${\stackrel{鈫�}{E}}_{net}=\stackrel{鈫�}{E_1}+\stackrel{鈫�}{E_2}+\stackrel{鈫�}{E_3}$

Hence, by substituting the values in the above equation, we get,

$$\begin{gathered} {\stackrel{鈫�}{E}}_{net}=8544.3\hat{i}+2848\hat{j}+10000\hat{i}+8544.3\hat{i}-2848\hat{j} \\ {\stackrel{鈫�}{E}}_{net}=27088.6\hat{i}N/C \end{gathered}$$

Hence, the net electrical field can be calculated as:

${\stackrel{鈫�}{E}}_{net}=27088.6\hat{i}N/C$