Question

In the region shown in Figure 13.63 there is an electric field due to a point charge located at the center of the dashed circle. The arrows indicate the magnitude and direction of the electric field at the locations shown

(a) What is the sign of the source charge? (b) Now a particle whose charge is $\mathbf{-}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{C}$is placed at location B. What is the direction of the electric force on the $\mathbf{-}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{C}$charge? (c) The electric field at location B has the value (2000,2000,0)N/C. What is the unit vector in the direction ofat this location? (d) What is the electric force on the $\mathbf{-}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{C}$charge? (e) What is the unit vector in the direction of this electric force?

Short answer

a) The sign of the source charge is negative.

b) The direction of force on $-7\times {10}^{-9}\mathrm{C}$is radially outwards.

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

d) the electric force exerted on charged particle is$\left(-1.4\times {10}^5,-1.4\times {10}^5,0\right)\mathrm{N}$ .

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

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of electric field line.

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.

(a) Determination of the sign of the source charge

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.

(b) Determination of the is the direction of the electric force on the-7×10-9C charge when another charge-7×10-9C is placed at location B.

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

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

(c) Determination of the unit vector in the direction E→ of at location B.

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

$\stackrel{⏜}{E}=\frac{\overrightarrow{E}}{\left|\overrightarrow{E}\right|}$

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

Substitute all the values in the above,

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

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

(d) Determination of the electric force on the -7×10-9C charge particle.

The electric force on the charge particle is given by,

$\overrightarrow{F}=\overrightarrow{E}q$

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

Substitute values in the above expression.

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

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

(e) Determination of the unit vector in the direction of this electric force.

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

$\stackrel{⏜}{F}=\frac{\overrightarrow{F}}{\left|\overrightarrow{F}\right|}$

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

Substitute values in the above,

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

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