Question

The electric field is a vector and so two electric fields at the same point must be added according to the laws of vector addition. Consider two charges top= $+ 2.00 μC$ and bottom $- 2.00 μC$ , separated by $a = 10 cm$ and a point $P$ at a distance $d = 30 cm$ as shown in the diagram. The diagram shows the direction of the electric fields produced at $P$ by each charge. Determine the magnitude and direction of net electric field at $P$

Accepted Answer

Given the value of each charge $q = \pm 2 \times 10^{- 6} N$

The distance between the charge $a = 5 cm = 5 \times 10^{- 2} m$

The distance of the point from the line joining the charge $d = 30 cm = 30 \times 10^{- 2} m$

The electric field will be working in the directions as shown. These are equal in magnitude.

If we resolve each of the electric fields along with the horizontal and vertical directions. Evidently, the magnitudes of the two electric fields (shown in green) are equal due to the geometry of the problem and so their horizontal components cancel each other and their vertical components add together.

The electric field due to each charge at $P$ =

$E = \frac{1}{4 {πε}_{0}} . \frac{q}{x^{2}}$ where $x$ = the distance between the charge and the point.

Substituting the values, we have

$E = \frac{1}{4 {πε}_{0}} . \frac{q}{r^{2} + a^{2}} = \frac{9 \times 10^{9} N^{2} m^{2} C^{- 1} \times 2.0 \times 10^{- 6}}{{(5^{2} + 30^{2})}^{2}} = \frac{18 \times 10^{3}}{925}$

The net electric field at $d = 30 cm$ in the vertical direction $E_{net} = 2 E \sin θ$

Where $θ$ is the semi vertex angle given by $\sin θ = \frac{\frac{a}{2}}{\sqrt{{(\frac{a}{2})}^{2} + d^{2}}} = \frac{5}{\sqrt{5^{2} + 30^{2}}} = \frac{5}{30.4}$

Therefore, we, have $E_{net} = \frac{2 \times 18 \times 10^{3}}{925} \times \frac{5}{30.4} = 6.4 \times 10^{4} {NC}^{- 1}$

The direction is vertically downwards

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