Exemplo 18.4 de College Physics, 18.5 Electric Field Lines: Multiple Charges

Adding Electric Fields

Problema e solução em inglês, como no livro original.

Find the magnitude and direction of the total electric field due to the two point charges, $q_{1}$ and $q_{2}$ , at the origin of the coordinate system as shown in Figure 18.24.

Two charges are placed on a coordinate axes. Q two is at the position x equals 4 and y equals 0 centimeters. Q one is at the position x equals 0 and y equals two centimeters. Charge on q one is plus five point zero nano coulomb and charge on q two is plus ten nano coulomb. The electric field, E one having a magnitude of one point one three multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow along positive y axis starting from the origin. The electric field, E two having magnitude zero point five six multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow along negative x axis starting from the origin. The resultant field makes an angle of sixty three point four degree above the negative y axis having magnitude one point two six multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow pointing away from the origin in the second quadrant.

Figure 18.24 The electric fields

E_{1}

and

E_{2}

at the origin O add to

E_{tot}

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia
Since the electric field is a vector (having magnitude and direction), we add electric fields with the same vector techniques used for other types of vectors. We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in this instance. We pretend that there is a positive test charge, $q$ , at point O, which allows us to determine the direction of the fields $E_{1}$ and $E_{2}$ . Once those fields are found, the total field can be determined using vector addition.
Solução
The electric field strength at the origin due to $q_{1}$ is labeled $E_{1}$ and is calculated:

$\begin{array}{l} E_{1} = k \frac{q_{1}}{r_{1}^{2}} = (8 . 99 \times 10^{9} N \cdot m^{2} {/C}^{2}) \frac{(5 . 00 \times 10^{- 9} C)}{{(2 . 00 \times 10^{- 2} m)}^{2}} \\ E_{1} = 1 . 124 \times 10^{5} N/C . \end{array}$
18.16

Similarly, $E_{2}$ is

$\begin{array}{l} E_{2} = k \frac{q_{2}}{r_{2}^{2}} = (8 . 99 \times 10^{9} N \cdot m^{2} {/C}^{2}) \frac{(10 . 0 \times 10^{- 9} C)}{{(4 . 00 \times 10^{- 2} m)}^{2}} \\ E_{2} = 0 . 5619 \times 10^{5} N/C . \end{array}$
18.17

Four digits have been retained in this solution to illustrate that $E_{1}$ is exactly twice the magnitude of $E_{2}$ . Now arrows are drawn to represent the magnitudes and directions of $E_{1}$ and $E_{2}$ . (See Figure 18.24.) The direction of the electric field is that of the force on a positive charge so both arrows point directly away from the positive charges that create them. The arrow for $E_{1}$ is exactly twice the length of that for $E_{2}$ . The arrows form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field $E_{tot}$ is
$\begin{array}{l} E_{tot} & = & (E_{1}^{2} + E_{2}^{2})^{1/2} \\ = & {(1.124 \times 10^{5} N/C)^{2} + (0.5619 \times 10^{5} N/C)^{2}}^{1/2} \\ = & 1.26 \times 10^{5} N/C. \end{array}$
18.18

The direction is

$\begin{array}{l} θ & = & {tan}^{- 1} (\frac{E_{1}}{E_{2}}) \\ = & {tan}^{- 1} (\frac{1 . 124 \times 10^{5} N/C}{0 . 5619 \times 10^{5} N/C}) \\ = & 63 . 4º, \end{array}$
18.19

or $63.4º$ above the x-axis.
Discussão
In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The total electric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges over an entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number of points in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.

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