Question

Calculate the electric field at the center of a square 52.5cm on a side if one corner is occupied by -38.6*10^-6 coloumb charge and the other three are occupied by -27.0*10^-6 coloumb charges.

Answer 1

Let's calculate the electric field in the center of the square, caused by each charge apart. Let's numerate corners from 1 to 4 and vectors of electric field, caused by charges in respective corners. (see picture)

Let's mark charges in the following way:

$q_1=q_3=q_4=\dfrac{-27.0}{10^6}$ C

$q_2=\dfrac{-38.6}{10^6}$ C

According to definition of electric field, vector of electric field is directed in the same direction as the force applied to +1 coulumb charge. Charges in all corners in this task are negative, to a positive charge would be attracted by them. Thus, the force, applied to +1 coulumb charge in the center of the square would be directed from the center to a corner. Thus, the electric field in the center of the square, caused by a charge is directed from the center to the charge.

The resultant electric field $\overrightarrow{E}$ in the center is the vector sum of all 4 electric fiels, caused by each charge apart.

$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}+\overrightarrow{E}_4$

The sum of $\overrightarrow{E_3}$ and $\overrightarrow{E_1}$ is equal to 0, because these vectors have equal magnitudes and opposite directions. $\overrightarrow{E_1}+\overrightarrow{E_3}=0$

Therefore, $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}+\overrightarrow{E_4}=0+\overrightarrow{E_2}+\overrightarrow{E_4}=\overrightarrow{E_2}+\overrightarrow{E_4}$

$\overrightarrow{E_2}$ and $\overrightarrow{E_4}$ have opposite directions, so the magnitude of their sum is equal to the diffterence of their magnitudes.

$|\overrightarrow{E}|=|\overrightarrow{E_2}|-|\overrightarrow{E_4}|$

Let's mark square's side lenght as $a$ and the distance from the center to the corner as $b$. It is known, that diagonals of a square are perpendicular. Thus, according to Pythagorean theorem, $b^2+b^2=a^2$

$2*b^2=a^2$ . $k$ is Coulumb's constant.

$b^2=\dfrac{a^2}{2}$

$|\overrightarrow{E_2}|=k*\dfrac{|q_2|}{b^2}=2k*\dfrac{|q_2|}{a^2}$

$|\overrightarrow{E_4}|=k*\dfrac{|q_4|}{b^2}=2k*\dfrac{|q_4|}{a^2}$

$|\overrightarrow{E}|=|\overrightarrow{E_2}|-|\overrightarrow{E_4}|=2k*\dfrac{|q_2|}{a^2}-2k*\dfrac{|q_4|}{a^2}=\dfrac{2k}{a^2}(|q_2|-|q_4|)$

$|\overrightarrow{E}|=\dfrac{2*9*10^9}{0.525^2}*(\dfrac{38.6}{10^6}-\dfrac{27.0}{10^6})=7.6*10^5 \dfrac{N}{C}$

Answer: $7.6*10^5 \dfrac{N}{C}$