Question

A charged particle located at the origin creates an electric field of $<-1.2\times {10}^3,0,0>\mathbf{}\mathbf{\text{N/C}}$at a location $<0.12,0,0>\mathbf{}\mathbf{\text{m}}$. What is the particle’s charge?

Short answer

The charge on the particle is $-1.9\times {10}^{-9}\text{C}$.

Step-by-step solution

Identification of given data 

The given data can be listed below,

• The electric field due to charged particle is, .
• The location of the electric field is, .

 Step 2: Concept/Significance of electric charge. 

When put in an electromagnetic field, charge is a physical feature of matter that allows it to feel force. Positive and negative charges are also possible.

Determination of the particle’s charge

The distance vector is given by,

The unit vector of distance is given by,

$\hat{r}=\frac{\overrightarrow{r}}{\left|\overrightarrow{r}\right|}$

Here, $\overrightarrow{r}$ is the distance vector and $\left|\overrightarrow{r}\right|$is the magnitude of the distance vector.

Substitute all the values in the above,

The electric field is given by,

$E=\frac{q}{4\pi {\epsilon }_0{r}^2}\hat{r}$

Here,q is the charge on the point charge,$\frac{1}{4\pi {\epsilon }_0}$ is the coulomb constant whose value is$9\times {10}^9\text{N}·{\text{m}}^2{\text{/C}}^2$ , r is the distance between observation location and source location and $\hat{r}$is the unit vector in direction of r.

Substitute all the values in the above,

Thus, the charge on the particle is $-1.9\times {10}^{-9}\text{C}$.