Question

Four charges 5µC, 5μC, -5µC, -5μC are placed at the corners of A, B, C, D of a square of side 1m. Calculate the electric field intensity at the center of the square.​

Answer 1

Electric field

• Direction is away from positive charge and towards the negative.
• It is a vector quantity.
• $\rm E = \dfrac{kq}{r^2}$

where , $\rm q = 5\times 10^{-6}, r =\dfrac{a}{\sqrt{2}}$

Solution

Refer the attached image resultant direction by every charge.

$\to\rm \vec{E_{net}} = \vec{E_A} + \vec{E_B} + \vec{E_C} +\vec{E_D}$

By adding the vectors we get;

$\to\rm E_{net} = \sqrt{ (2E)^2 +(2E)^2 + 2(E)(E)cos90\degree}$

$\to\rm E_{net} = \sqrt{ (2E)^2.(1+1)+0}$

$\to\sf E_{net} = 2\sqrt{2} \: E$

$\to\rm E_{net} = 2\sqrt{2} \: \times \dfrac{k\times 5\times 10^{-6}}{\bigg(\dfrac{a}{\sqrt{2}}\bigg)^2}$

$\to\rm E_{net} = 2\sqrt{2} \: \times \dfrac{9\times 10^9\times 5\times 10^{-6}}{\bigg(\dfrac{a^2}{2}\bigg)}$

$\to\rm E_{net} = 2\sqrt{2} \: \times \dfrac{45\times 10^3\times 2}{1^2}$

$\to\rm E_{net} = 2\sqrt{2} \: \times 90\times 10^3$

$\to\rm E_{net} = 18\sqrt{2} \times 10^4$

$\implies\bf E_{center} = 18\sqrt{2} \times 10^4 \: \hat{-j}$

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Answer 2

Explanation:

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿Yøur-Añswer♡}}}$

$\large\bf{\underline{\red{VERIFIED✔}}}$

Electric field

Direction is away from positive charge and towards the negative.

It is a vector quantity.

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

where , $\rm q = 5\times 10^{-6}, r =\dfrac{a}{\sqrt{2}}$

Solution

Refer the attached image resultant direction by every charge.

$\to\rm \vec{E_{net}} = \vec{E_A} + \vec{E_B} + \vec{E_C} +\vec{E_D}$

By adding the vectors we get;

$\to\rm E_{net} = \sqrt{ (2E)^2 +(2E)^2 + 2(E)(E)cos90\degree}$

$\to\rm E_{net} = \sqrt{ (2E)^2.(1+1)+0}$

$\to\sf E_{net} = 2\sqrt{2} \: E$

$\to\rm E_{net} = 2\sqrt{2} \: \times \dfrac{k\times 5\times 10^{-6}}{\bigg(\dfrac{a}{\sqrt{2}}\bigg)^2}$

$\to\rm E_{net} = 2\sqrt{2} \: \times \dfrac{9\times 10^9\times 5\times 10^{-6}}{\bigg(\dfrac{a^2}{2}\bigg)}$

$\to\rm E_{net} = 2\sqrt{2} \: \times \dfrac{45\times 10^3\times 2}{1^2}$

$\to\rm E_{net} = 2\sqrt{2} \: \times 90\times 10^3$

$\to\rm E_{net} = 18\sqrt{2} \times 10^4$

$\implies\bf E_{center} = 18\sqrt{2} \times 10^4 \: \hat{-j}$

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