Respuesta :
Answer:
[tex]\frac{E_{A}}{E_{B}}=4[/tex]
Explanation:
The electric field is defined as the electric force per unit of charge, this is:
[tex]E=\frac{F}{q}[/tex].
The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as
[tex]F=\frac{kQq}{r^{2}}[/tex].
By substitution we get that
[tex]E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}[/tex]
Now, letting [tex]E_{A}[/tex] be the electric field at point A, letting [tex]E_{B}[/tex] be the electric field at point B, and letting R be the distance from the charge to A:
[tex]E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}[/tex].
The ration of the electric fields is
[tex]\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4[/tex]
This means that at half the distance, the electric field is four times stronger.
The ratio of the electric field strength at point A to the electric field strength at point B is 4.
Given to us
A positive charge Q and a point B is twice as far away from Q as point A.
What is Coulomb's Law?
According to Coulomb's law, the electric force between two electrical charges particles is inversely proportional to the square of the distance between them and it is directly proportional to the product of their charges.
[tex]F =\dfrac{k\ Q_1 Q_2}{r^2}[/tex]
We know that an electric field is written as,
[tex]F= EQ[/tex]
Merging the two-equation we get
[tex]E = \dfrac{k Q}{r^2}[/tex]
What is the ratio of the electric field?
As we can write
[tex]\dfrac{E_A}{E_B} = \dfrac{\dfrac{k Q}{R^2}}{\dfrac{k Q}{(2R)^2}}[/tex]
[tex]\dfrac{E_A}{E_B} = 4[/tex]
Hence, the ratio of the electric field strength at point A to the electric field strength at point B is 4.
Learn more about Coulomb's Law:
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