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Point charges of 16.7 nC and 47.3 nC are placed 0.500 m apart. What is the electric field halfway between them? Indicate direction by a positive or a negative value. Keep in mind that a positive vector is one directed to the right and a negative vector is one directed to the left. Your answer should be a positive or a negative number with two decimal places, do not include the unit. Hint: 1 nC = 10-9 C

Respuesta :

Answer:

-4320

Explanation:

Using the formula for electric field between two points

E = kq/r²

The electric field at a midpoint, say x, is

E(x) = [(k * 16.7*10^-9)/0.5²] - [(k * 47.3*10^-9)/0.25²]

E(x) = [k * (16.7*10^-9 - 47.3*10^-9)/0.25²]

E(x) = [(k * -30.6*10^-9)/0.0625]

E(x) = [(9*10^-9 * -30.6*10^-9)/0.0625]

E(x) = -270/0.0625

E(x) = -4320

Therefore, we can conclude that the electric field, halfway in between them is -4320

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