The vector characteristics of the electric field allow to find the result for the total electric field in the center of the square is:
- Total electric field is: E = 7.05 10⁵ C/m
The electric field is a vector quantity, given by the ratio of the electric force on a test charge and divided by the value of the test charge.
[tex]E = k \frac{q}{r^2}[/tex]
Where k is the Coulomb constant, q the point charge and r the distance from the charge to the point of interest.
The electric field in the inside square is the vector sum of the fields created by each charge.
E[tex]_{total}[/tex] = E₁ + E₂ + E₃ + E₄
In the attached we can see a diagram of the distribution of the charges, where q₁ = -38.0 10⁻⁶ C and the other charges are:
q₂ = q₃ = q₄ = -27.2 10⁻⁶ C
The distance from the corner to the center of the square is:
D = [tex]\sqrt{a^2 + a^2} = \sqrt{2} \ a[/tex]
The distance to the midpoint of the diagonal is:
[tex]d = \frac{D}{2} \\d = \frac{a}{\sqrt{2} } \\\\d= \frac{0.525}{\sqrt{2} }[/tex]
d = 0.3712 m
In the adjoint we can see that the electric field for charges 2 and 4 is directed in the same diagonal and in the opposite direction, therefore its sum gives zero.
The electric field of charges q₁ and q₃ are on the same line, so we can perform the algebraic sum.
[tex]E_{total} = - E_1 + E_2 \\E_{total} = \frac{k}{d^2} \ (-q_1 + q_2)[/tex]
Let's calculate the electric field.
[tex]E_{total} =\frac{9 \ 10^9}{0.3715^2 } \ ( - 38.0 + 27.2 ) \ 10^{-6} \\E_{total}= - 7.05 \ 10^5 C/m[/tex]
In conclusion using the vector characteristics of the electric field we can find the result for the total electric field in the center of the square is:
- Total electric field E = 7.05 10⁵ C / m
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