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three 3.0 -μc point charges are at the three corners of a square of side 0.50 m. the last corner is occupied by a −3.0 -μc charge. find the magnitude of the electric field at the center of the square.

Respuesta :

The magnitude of the electric field at the center of the square is 4.32 * [tex]10^{5}[/tex] N / C

We know that,

The magnitude of the electric field by the one positive charge and other negative charge at the centre is

E = k ( [tex]q_{2}[/tex] - [tex]q_{4}[/tex] ) / [tex]r^{2}[/tex]

The distance of each charge from the centre of the square is

r = d / √ 2

where,

k = Coulomb's constant

[tex]q_{1}[/tex], [tex]q_{2}[/tex], [tex]q_{3}[/tex], [tex]q_{4}[/tex] = Point charges

d = Side of square

Given that,

[tex]q_{1}[/tex] = [tex]q_{2}[/tex] = [tex]q_{3}[/tex] = 3 μ C = 3 * [tex]10^{-6}[/tex] C

[tex]q_{4}[/tex] = - 3 μ C = - 3 * [tex]10^{-6}[/tex] C

d = 0.5 m

k = 9 * [tex]10^{9}[/tex] N m² / C²

E = k ( [tex]q_{2}[/tex] - [tex]q_{4}[/tex] ) / [tex]r^{2}[/tex]

E = 2 k / d² ( [tex]q_{2}[/tex] - [tex]q_{4}[/tex] )

E = 2 ( 9 * [tex]10^{9}[/tex] ) ( 3 * [tex]10^{-6}[/tex] - ( - 3 * [tex]10^{-6}[/tex] )) / 0.5²

E = 4.32 * [tex]10^{5}[/tex] N / C

Electric field is the field surrounding an electrically charged particle. If any other charged particle enters the field, it exerts force on the particle.

Therefore, the magnitude of the electric field at the center of the square is 4.32 * [tex]10^{5}[/tex] N / C

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