Irodov 3.6: The Field Vector in Components – Amit & Deepali Sinha Foundation

Problem Irodov 3.6

A positive point charge of $50$ $\mu$ C lies in the $xy$ plane at the point with position vector $\vect{r}_0 = 2\vect{i} + 3\vect{j}$ , where $\vect{i}$ and $\vect{j}$ are the unit vectors along the $x$ and $y$ axes and distances are in metres. Find the electric field vector $\vect{E}$ and its magnitude at the point with position vector $\vect{r} = 8\vect{i} - 5\vect{j}$ .

The charge q at r_0 = (2, 3) m and the field point P at r = (8, -5) m, drawn to scale. — The charge $q$ at $\vect{r}_0 = (2, 3)$ m and the field point $P$ at $\vect{r} = (8, -5)$ m, drawn to scale.

Given

$q = 50$ $\mu$ C
$\vect{r}_0 = 2\vect{i} + 3\vect{j}$ m
$\vect{r} = 8\vect{i} - 5\vect{j}$ m

Find

the vector

\vect{E}

\vect{r}

and its magnitude

E

Solution

Step 1Coulomb’s law in vector form

The field of a point charge $q$ at $\vect{r}_0$ points along the separation vector $\vect{r} - \vect{r}_0$ from the charge to the field point, and falls off as the inverse square of its length:

\vect{E} = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,(\vect{r} - \vect{r}_0)}{\lvert\vect{r} - \vect{r}_0\rvert^3} \tag{1}

Step 2The separation vector

Subtracting the two position vectors component by component, then taking the length:

\begin{aligned} \vect{r} - \vect{r}_0 &= (8 - 2)\,\vect{i} + (-5 - 3)\,\vect{j} = 6\,\vect{i} - 8\,\vect{j}\ \text{m}\\ \lvert\vect{r} - \vect{r}_0\rvert &= \sqrt{6^2 + 8^2} = 10\ \text{m} \end{aligned}

The separation vector r - r_0 from the charge to P, and the field E at P pointing along it, away from the positive charge. — The separation vector $\vect{r} - \vect{r}_0$ from the charge to $P$ , and the field $\vect{E}$ at $P$ pointing along it, away from the positive charge.

Step 3The field vector and its magnitude

Substituting into (1), with $1/(4\pi\varepsilon_0) = 9.0\times10^{9}$ m/F:

\begin{aligned} \vect{E} &= (9.0\times10^{9})\,\frac{(50\times10^{-6})\,(6\,\vect{i} - 8\,\vect{j})}{10^3}\\ &= 450\,(6\,\vect{i} - 8\,\vect{j})\ \text{V/m} \end{aligned}

\vect{E} = (2.7\,\vect{i} - 3.6\,\vect{j})\ \text{kV/m}

(2)

The magnitude is the factor $450$ V/m times the length $10$ of $6\,\vect{i} - 8\,\vect{j}$ :

E = 450\times 10 = 4.5\times10^{3}\ \text{V/m}

In symbols, this is Coulomb’s law with the separation found above:

E = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{\lvert\vect{r} - \vect{r}_0\rvert^2} = 4.5\ \text{kV/m}

(3)

Check

The components give the magnitude independently:

\sqrt{2.7^2 + 3.6^2} = \sqrt{7.29 + 12.96} = \sqrt{20.25} = 4.5\ \text{kV/m}

Direction: $E_x > 0$ and $E_y < 0$ , and $P$ lies to the right of and below the charge, so the field of the positive charge points away from it, as it must.

Remark

Only the difference $\vect{r} - \vect{r}_0$ enters (1). Moving the origin shifts both position vectors by the same amount and leaves $\vect{E}$ unchanged, which is why the vector form is the natural starting point for superposition: the field of many charges is the sum of such terms, one per charge, each built from its own separation vector.