Respuesta :
Answer:
Explanation:
Electric field due to charge at origin
= k Q / r²
k is a constant , Q is charge and r is distance
= 9 x 10⁹ x 5 x 10⁻⁶ / .5²
= 180 x 10³ N /C
In vector form
E₁ = 180 x 10³ j
Electric field due to q₂ charge
= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²
= 30.33 x 10³ N / C
It will have negative slope θ with x axis
Tan θ = .5 / √.5² + .8²
= .5 / .94
θ = 28°
E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j
= 26.78 x 10³ i - 14.24 x 10³ j
Total electric field
E = E₁ + E₂
= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j
= 26.78 x 10³ i + 165.76 X 10³ j
magnitude
= √(26.78² + 165.76² ) x 10³ N /C
= 167.8 x 10³ N / C .
The electric field at a point 0.5 m away from the origin is [tex]167.8 \times 10^2\rm \ N / C .[/tex]
What is an electric field?
The area around a charged particle can affect another charged particle either as attraction or repulsion.
Electric field due to charges at the origin
[tex]E = \dfrac {k Q }{ r^2}[/tex]
Where,
k - constant ,
Q - charge
r - distance
Put the value in the formula,
[tex]E= \dfrac {9 \times 10^9 \times 5 \times 10^{-6} }{5^2}\\\\E= 180 \times 10^3 \rm \ N /C[/tex]
In vector form
[tex]E_1 = 180 \times 10^3 j[/tex]
Electric field due to q₂ charge,
[tex]E_2= \dfrac {9 \times 10^9 \times 3 \times 10^{-6} }{0.5^2\times 0.8 ^2}\\\\E_2= 30.33\times 10^3 \rm \ N /C[/tex]
This will have a negative slope [tex]\theta[/tex] with x-axis
So,
[tex]tan \theta = \dfrac {0.5} { \sqrt {0.5^2 + 0.8^2}}\\\\tan \theta = \dfrac {0.5} {0.94}\\\\\theta = 28^o[/tex]
Thus,
[tex]E_2 = 30.33 \times 10^3 cos 28 i - 30.33 \times 10^3 sin28j\\\\E_2 = 26.78 \times 10^3 i - 14.24 \times 10^3 j[/tex]
The magnitude of the total electric field
[tex]E = E_1 + E_2E= \sqrt {(26.78^2 + 165.76^2 ) \times 10^2 \rm \ N /C}E= 167.8 x 10^2\rm \ N / C .[/tex]
Therefore, the electric field at a point 0.5 m away from the origin is [tex]167.8 \times 10^2\rm \ N / C .[/tex]
To know more about point charge,
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