Respuesta :
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
[tex]E = \frac{Kq}{r^2}[/tex]
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 = [tex]\sqrt{2}\times a[/tex]
where, a be the side of the square
So, Electric field at B due to charge at A
[tex]E_{A}=\frac{Kq}{a^2}[/tex]
[tex]E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]
EA = 6531.32 N/C
Electric field at B due to charge at C
[tex]E_{C}=\frac{Kq}{a^2}[/tex]
[tex]E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]
Ec = 6531.32 N/C
Electric field at B due to charge at D
[tex]E_{D}=\frac{Kq}{2a^2}[/tex]
[tex]E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}[/tex]
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
[tex]E=\sqrt{E_{x}^{2}+E_{y}^{2}}[/tex]
[tex]E=\sqrt{8840.5^{2}+8840.5^{2}}[/tex]
E = 12500 N/C
Explanation:
Answer:
[tex]1.250\times 10^4\ N/C.[/tex]
Explanation:
Given:
- Charge on each corner of the square, [tex]q=4.25\times 10^{-6}\ C.[/tex]
- Length of the side of the square, [tex]a = 2.42\ m.[/tex]
According to the Coulomb's law, the strength of the electric field at a point due to a charge [tex]q[/tex] at a point [tex]r[/tex] distance away is given by
[tex] E = \dfrac{kq}{r^2}[/tex]
where,
[tex]k[/tex] = Coulomb's constant = [tex]9\times 10^9\ Nm^2/C^2[/tex].
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
[tex]E_A = \dfrac{kq}{a^2}[/tex]
Since, this electric field is along positive x axis direction, therefore,
[tex]\vec E_A = \dfrac{kq}{a^2}\ \hat i.[/tex]
[tex]\hat i[/tex] 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
[tex]E_B = \dfrac{kq}{a^2}[/tex]
Since, this electric field is along negative y axis direction, therefore,
[tex]\vec E_B = \dfrac{kq}{a^2}\ (-\hat j).[/tex]
[tex]\hat j[/tex] 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
[tex]E_C = \dfrac{kq}{r^2}[/tex]
where, [tex]r=\sqrt{a^2+a^2}=a\sqrt 2[/tex].
Since, this electric field is along the direction, which is making an angle of [tex]45^\circ[/tex] below the positive x-axis direction, therefore, the direction of this electric field is given by [tex]\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j)[/tex].
[tex]\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).\\[/tex]
Thus, the total electric field at the point P is given by
[tex]\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)[/tex]
[tex]=\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.[/tex]
The magnitude of the electric field at the given point due to all the three charges is given by
[tex]E=\sqrt{(8.837\times 10^3)^2+(-8.837\times 10^3)^2}=1.250\times 10^4\ N/C.[/tex]
