Question

four point charges are placed at the corners of a square as shown in the figure. each side of the square has a length 2.0 m. determine the magnitude of the electric field at the point p, the center of the square. +4 μc, +1 μc, -3 μc, +4 μc. $vec{e}=\frac{k|q|}{r^{2}}$

Explanation:

Step1: Calculate the distance from corner to center

For a square of side - length $a = 2.0$ m, the distance $r$ from a corner to the center of the square can be found using the Pythagorean theorem. The diagonal of the square $d=\sqrt{a^{2}+a^{2}}=\sqrt{2a^{2}}=\sqrt{2}a$, and $r = \frac{d}{2}=\frac{\sqrt{2}a}{2}$. Substituting $a = 2.0$ m, we get $r=\frac{\sqrt{2}\times2}{2}=\sqrt{2}$ m.

Step2: Calculate the electric - field due to each charge

The electric - field due to a point charge $q$ at a distance $r$ is given by $E=\frac{k|q|}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$.
For $q_1=4\times10^{- 6}\ C$, $E_1=\frac{k|q_1|}{r^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{(\sqrt{2})^{2}}=1.8\times10^{4}\ N/C$.
For $q_2=-3\times10^{-6}\ C$, $E_2=\frac{k|q_2|}{r^{2}}=\frac{9\times10^{9}\times3\times10^{-6}}{(\sqrt{2})^{2}} = 1.35\times10^{4}\ N/C$.
For $q_3 = 4\times10^{-6}\ C$, $E_3=\frac{k|q_3|}{r^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{(\sqrt{2})^{2}}=1.8\times10^{4}\ N/C$.
For $q_4 = 1\times10^{-6}\ C$, $E_4=\frac{k|q_4|}{r^{2}}=\frac{9\times10^{9}\times1\times10^{-6}}{(\sqrt{2})^{2}}=4.5\times10^{3}\ N/C$.

Step3: Resolve the electric - fields into components

Due to the symmetry of the square, we can consider the electric - fields in the $x$ and $y$ directions. Let's assume the positive $x$ - axis is to the right and the positive $y$ - axis is up.
The electric - fields from charges at opposite corners will have components that add up or cancel out.
The net electric - field in the $x$ - direction:
$E_x=(E_1 - E_2)\cos45^{\circ}+(E_3 - E_4)\cos45^{\circ}$
$E_1 - E_2=1.8\times10^{4}-1.35\times10^{4}=4.5\times10^{3}\ N/C$
$E_3 - E_4=1.8\times10^{4}-4.5\times10^{3}=1.35\times10^{4}\ N/C$
$E_x=(4.5\times10^{3}+1.35\times10^{4})\times\frac{\sqrt{2}}{2}=(18\times10^{3})\times\frac{\sqrt{2}}{2}=9\sqrt{2}\times10^{3}\ N/C$
The net electric - field in the $y$ - direction:
$E_y=(E_1 - E_2)\sin45^{\circ}+(E_3 - E_4)\sin45^{\circ}=(4.5\times10^{3}+1.35\times10^{4})\times\frac{\sqrt{2}}{2}=9\sqrt{2}\times10^{3}\ N/C$

Step4: Calculate the magnitude of the net electric - field

The magnitude of the net electric - field $E=\sqrt{E_x^{2}+E_y^{2}}$. Since $E_x = E_y=9\sqrt{2}\times10^{3}\ N/C$, $E=\sqrt{(9\sqrt{2}\times10^{3})^{2}+(9\sqrt{2}\times10^{3})^{2}}=\sqrt{2\times(9\sqrt{2}\times10^{3})^{2}}=1.8\times10^{4}\ N/C$

Answer:

$1.8\times10^{4}\ N/C$