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A positive test charge of 5.0 x 10^-4 C is in an electric field that exerts a force of 2.5 x 10^-4 N on it. What is the magnitude of the electric field at the location of the test charge?

Respuesta :

mavila18

Answer:

E = 0.5N/C

Explanation:

In order to calculate the magnitude of the electric field you use the following formula:

[tex]E=\frac{F}{q}[/tex]

q: charge = 5.0*10^-4 C

F: force on the charge = 2.5*10^-4N

You replace the values of q and F in the equation for E:

[tex]E=\frac{2.5*10^{-4}N}{5.0*10^{-4}C}=0.5\frac{N}{C}[/tex]

hence, the magnitude of the electric field at the position of the carge is 0.5N/C

Cricetus

At the location of the test charge, the magnitude will be:

"0.5 N/C".

Electric field

Whenever charge seems to be available inside any form, an electric property has been linked among each spatial position. This same size and direction are represented either by the quantity of E.

According to the question,

Charge, q = 5.0 × 10⁻⁴ C

Force on the charge, F = 2.5 × 10⁻⁴ N

We know the relation of the magnitude of electric field be

→ E = [tex]\frac{F}{q}[/tex]

By substituting the values, we get

     = [tex]\frac{2.5\times 10^{-4}}{5.0\times 10^{-4}}[/tex]

     = 0.5 N/C

Thus the approach above is correct.

Find out more information about electric field here:

https://brainly.com/question/14372859

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