91影视

The given data can be listed below,

• Charge present at location B is,$q=-7脳{10}^{-9}C$ .
• The value of electric field is,$\stackrel{鈫�}{E}=(2000,2000,0)\mathrm{N}/\mathrm{C}$

When a positive unit charge is placed in an electric field, it travels along electric field lines. The size of a charge is related to the number of electric field lines leaving or entering it.

As shown in the diagram the electric field lines are going inwards where the source charge is placed. Field lines always go from positive to negative direction so source must have negative charge on it.

Thus, the sign of source charge is negative.

When a like charge is present at location B the force exerted on charge$-7脳{10}^{-9}C$ have the direction radially outwards because two similar charges repel each other.

Thus, the direction of force on$-7脳{10}^{-9}C$ is radially outwards.

The unit vector in the direction of electric field is given by,

$\stackrel{铃�}{E}=\frac{\stackrel{鈫�}{E}}{\left|\stackrel{鈫�}{E}\right|}$

Here, $\stackrel{鈫�}{E}$is the electric field vector and $\left|\stackrel{鈫�}{E}\right|$is the magnitude of electric field.

Substitute all the values in the above,

$$\begin{gathered} \stackrel{铃�}{E}=\frac{\left(2000,2000,0\right)N/C}{\sqrt{{\left(2000N/C\right)}^2+{\left(2000N/C\right)}^2+0}} \\ =\frac{\left(2000,2000,0\right)}{2000\sqrt{2}} \\ =\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right) \end{gathered}$$

Thus, the value of unit vector in the direction of electric field is $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$.

The electric force on the charge particle is given by,

$\stackrel{鈫�}{F}=\stackrel{鈫�}{E}q$

Here,$\stackrel{鈫�}{E}$ is the electric field vector and q is the charge on the particle.

Substitute values in the above expression.

$$\begin{gathered} \stackrel{鈫�}{F}=\left(-7脳{10}^{-9}C\right)\left(2000,2000,0\right)\mathrm{N}/\mathrm{C} \\ =\left(-1.4脳{10}^{-5},-1.4脳{10}^{-5}\right)\mathrm{N} \end{gathered}$$

Thus, the electric force exerted on charged particle is$\left(-1.4脳{10}^{-5},-1.4脳{10}^{-5}\right)\mathrm{N}$ .

The unit vector in the direction of electric force is given by,

$\stackrel{铃�}{F}=\frac{\stackrel{鈫�}{F}}{\left|\stackrel{鈫�}{F}\right|}$

Here,$\stackrel{鈫�}{F}$is the force vector, andlocalid="1656918407941" $\left|\stackrel{鈫�}{E}\right|$is the magnitude of force vector.

Substitute values in the above,

localid="1656918698142" $$\begin{gathered} \stackrel{铃�}{F}=\frac{\left(-1.4脳{10}^{-5},-1.4脳{10}^{-5},0\right)N/C}{\sqrt{{\left(-1.4脳{10}^{-5}N\right)}^2+{\left(-1.4脳{10}^{-5}N\right)}^2+0}} \\ =\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right) \end{gathered}$$

Thus, the value of unit vector in the direction of electric force islocalid="1656918690412" $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$.