Question

Find the electric field at the center of square. Assume that q1=11.8nC, q2=-11.8nC, q3=23.6nC, q4=-23.6nC and a=5.2cm. Such that 'a' is the distance between any two charges. The charge q1 and q3 are along two sides of a diagonal.

Answer 1

Answer:

$E_T=[-27739.6\hat{i}-55479\hat{j}]\frac{N}{C}$

Explanation:

You have four charges at the corners of a square of side a=5.2cm

In order to calculate the electric field at the center of the square, you sum the contribution of the electric field generated by each charge.

The total electric field is given by:

$E_T=E_1+E_2+E_3+E_4$    (1)

each contribution to the total electric field has two components x and y. The signs of the components depends of the direction of the field, which is given by the sign of the charge that produced the electric field. Then, you have

$E_1=k\frac{q_1}{r^2}cos\theta\hat{i}-k\frac{q_1}{r^2}sin\theta\hat{j}\\\\E_1=k\frac{q_1}{r^2}(cos\theta \hat{i}-sin\theta \hat{j})$   (2)

q1 = 11.8*10^-9 C

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

For a square you obtain that

$r=\sqrt{2}a=\sqrt{2}(5.2cm)=7.35cm=7.35*10^{-2}m$

and the angle is 45°

Then, you have in the equation (2):

$E_1=(8.98*10^9Nm^2/C^2)\frac{11.8*10^{-9}C}{(7.35*10^{-2}m)^2}(cos45\° \hat{i}-sin45\° \hat{j})=[13869.7\hat{i}-13869.7\hat{j}]\frac{N}{C}$

In the same way you obtain for the other contributions to the total electric field:

For E2:

$E_2=k\frac{q_2}{r^2}(cos45\°\hat{i}+sin45\° \hat{j})\\\\E_2=[13869.7\hat{i}+13869.7\hat{j}]\frac{N}{C}$

For E3:

$E_3=k\frac{q_3}{r^2}(-cos45\°\hat{i}+sin45\°\hat{j})\\\\E_3=(8.98*10^9Nm2/C^2)\frac{23.6*10^{-9}C}{(7.35*10^{-2}m)^2}(-cos45\°\hat{i}+sin45\°\hat{j})\\\\E_3=39229.58(-cos45\°\hat{i}+sin45\°\hat{j})\frac{N}{C}\\\\E_3=[-27739.5\hat{i}+-27739.5\hat{j}]\frac{N}{C}$

for E4:

$E_4=k\frac{q_4}{r^2}(-cos45\°\hat{i}-sin45\°\hat{j})\\\\E_4=[-27739.5\hat{i}-27739.5\hat{j}]\frac{N}{C}$

Finally, you sum component by component the four contributions to the total electric field (equation (1)):

$E_T=[-27739.6\hat{i}-55479\hat{j}]\frac{N}{C}$

Answer 2

There still are four chargers in the corners of a square with side a=5.2cm. To compute the electric field just at the center of the composition, add the contributions of each shield's electromagnetic current.

entire electric field

$\to E_T=E_1+E_2+E_3+E_4\ \ \ \ \ \ \ (1)$

Every contribution to the total electric field is comprised of two components, x, and y. These components' signs were determined by the direction of the field, which is determined by the sign of the charge that produced the electric field. There's

$\to E_1=k\frac{q_1}{r^2}\cos\theta\hat{i}-k\frac{q_1}{r^2}\sin\theta\hat{j}\\\\\to E_1=k\frac{q_1}{r^2}(\cos\theta \hat{i}-\sin\theta \hat{j}) \ \ \ \ (2)\\\\\to q_1 = 11.8\times 10^{-9}\ C$

Coulomb’s constant (K) $= 8.98\times 10^9\ \frac{Nm^2}{C^2}$

square, you obtain that

$\to r=\sqrt{2}\\\\\to a=\sqrt{2}(5.2\ cm)=7.35\ cm=7.35\times 10^{-2}\ m$

angle= $45\°$  then, you have in the equation (2):

$\to E_1=(8.98\times 10^9\ \frac{Nm^2}{C^2})\frac{11.8 \times 10^{-9}\ C}{(7.35 \times 10^{-2}m)^2}(\cos 45^{\circ} \hat{i}-\sin 45^{\circ} \hat{j})$

         $=[13869.7\hat{i}-13869.7\hat{j}]\ \frac{N}{C}$

Similarly, for other components to the overall electric field, you get:

For E2:

$\to E_2=k\frac{q_2}{r^2}(\cos45^{\circ}\hat{i}+\sin45^{\circ} \hat{j})\\\\ \to E_2=[13869.7\hat{i}+13869.7\hat{j}]\ \frac{N}{C}$

For E3:

$\to E_3=k\frac{q_3}{r^2}(-\cos45^{\circ}\hat{i}+\sin45^{\circ} \hat{j})\\\\\to E_3=(8.98\times 10^9 \ \frac{Nm^2}{C^2}) \ \frac{23.6 \times 10^{-9}\ C}{(7.35\times 10^{-2}\ m)^2}(-\cos45^{\circ} \hat{i}+\sin45^{\circ} \hat{j})\\\\\to E_3=39229.58(-\cos45^{\circ} \hat{i}+\sin45^{\circ} \hat{j})\frac{N}{C}\\\\\to E_3=[-27739.5\hat{i}+-27739.5\hat{j}]\ \frac{N}{C}$

For E4:  

$\to E_4=k\frac{q_4}{r^2}(-\cos45^{\circ}\hat{i}-\sin45^{\circ}\hat{j})\\\\ \to E_4=[-27739.5\hat{i}-27739.5\hat{j}]\ \frac{N}{C}$

Finally, we add up the four contributions to the total electric field (equation (1)):

$E_T=[-27739.6\hat{i}-55479\hat{j}]\frac{N}{C}$

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