Respuesta :
Answer:
Electric field, [tex]E=4\times 10^5\ N/C[/tex]
Explanation:
It is given that,
Magnitude of charge, [tex]q_o=2\ nC=2\times 10^{-9}\ C[/tex]
Force experienced, [tex]F=8\times 10^{-4}\ N[/tex]
We need to find the electric field at the origin. It is given by :
[tex]F=q_o\times E[/tex]
[tex]E=\dfrac{F}{q_o}[/tex]
[tex]E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}[/tex]
[tex]E=4\times 10^5\ N/C[/tex]
So, the electric field at the origin is [tex]4\times 10^5\ N/C[/tex]. Hence, this is the required solution.
The electric field at the origin by the test charge is 4 x 10⁵ N/C.
What is electric field?
The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.
Given is a test charge q0 = 2 nC = 2 x 10⁻⁹ C is placed at the origin, it experiences a force of 8 x 10⁻⁴ N in the positive y direction.
Force experienced by the charge,
F = qE
Substitute the values, we get
8 x 10⁻⁴ = 2 x 10⁻⁹ x E
E = 4 x 10⁵ N/C
Hence, the electric field at the origin is 4 x 10⁵ N/C.
Learn more about electric field.
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