Question

Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied by 4.25×10^−6 C charges. Express your answer to three significant figures and include the appropriate units.

Answer 1

Ans:

12500 N/C

Explanation:

Side of square,  a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

$E = \frac{Kq}{r^2}$

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD = $\sqrt{2}\times a$

where, a be the side of the square

So, Electric field at B due to charge at A

$E_{A}=\frac{Kq}{a^2}$

$E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}$

EA = 6531.32 N/C

Electric field at B due to charge at C

$E_{C}=\frac{Kq}{a^2}$

$E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}$

Ec = 6531.32 N/C

Electric field at B due to charge at D

$E_{D}=\frac{Kq}{2a^2}$

$E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}$

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

$E=\sqrt{E_{x}^{2}+E_{y}^{2}}$

$E=\sqrt{8840.5^{2}+8840.5^{2}}$

E = 12500 N/C

Explanation:

Answer 2

Answer:

$1.250\times 10^4\ N/C.$

Explanation:

Given:

• Charge on each corner of the square, $q=4.25\times 10^{-6}\ C.$
• Length of the side of the square, $a = 2.42\ m.$

According to the Coulomb's law, the strength of the electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$E = \dfrac{kq}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ Nm^2/C^2$.

The direction of the electric field is along the line joining the point an d the charge.

The electric field at the point P due to charge at A is given by

$E_A = \dfrac{kq}{a^2}$

Since, this electric field is along positive x axis direction, therefore,

$\vec E_A = \dfrac{kq}{a^2}\ \hat i.$

$\hat i$ is the unit vector along the positive x-axis direction.

The electric field at the point P due to charge at B is given by

$E_B = \dfrac{kq}{a^2}$

Since, this electric field is along negative y axis direction, therefore,

$\vec E_B = \dfrac{kq}{a^2}\ (-\hat j).$

$\hat j$ is the unit vector along the positive y-axis direction.

The electric field at the point P due to charge at C is given by

$E_C = \dfrac{kq}{r^2}$

where, $r=\sqrt{a^2+a^2}=a\sqrt 2$.

Since, this electric field is along the direction, which is making an angle of $45^\circ$ below the positive x-axis direction, therefore, the direction of this electric field is given by $\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j)$.

$\vec E_C = \dfrac{kq}{2a^2}\ (\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j))\\=\dfrac{kq}{2a^2}\ (\dfrac{1}{\sqrt 2}\hat i-\dfrac{1}{\sqrt 2}\hat j)\\=\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).$

Thus, the total electric field at the point P is given by

$\vec E = \vec E_A+\vec E_B +\vec E_C\\=\dfrac{kq}{a^2}\hat i+\dfrac{kq}{a^2}(-\hat j)+\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\=\left ( \dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )\hat i+\left (\dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )(-\hat j)\\=\dfrac{kq}{a^2}\left [\left ( 1+\dfrac{1}{2\sqrt 2} \right )\hat i+\left (1+\dfrac{1}{2\sqrt 2} \right )(-\hat j)\right ]\\=\dfrac{kq}{a^2}(1.353\hat i-1.353\hat j)$

$=\dfrac{(9\times 10^9)\times (4.25\times 10^{-6})}{2.42^2}\times (1.353\hat i-1.353\hat j)\\=(8.837\times 10^3\hat i-8.837\times 10^3\hat j)\ N/C.$

The magnitude of the electric field at the given point due to all the three charges is given by

$E=\sqrt{(8.837\times 10^3)^2+(-8.837\times 10^3)^2}=1.250\times 10^4\ N/C.$