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What is the magnitude and direction of an electric field that exerts a 2.00×10−5N upward force on a –1.75μC charge?

Respuesta :

asuwafohjames

The magnitude and direction of the electric field is  -14.3 N/C  Downward.

To calculate the electric field, we use the formula below.

Formula:

  • E = F/q................ Equation 1

Where:

  • E = Electric Field,
  • F = Force exerted by the field
  • q = Electric charge.

From the question below,

Given:

  • F = 2.5×10⁻⁵ N
  • q = 1.75 μC = -1.75×10⁻⁶ C

Substitute these values into equation 1

  • E = (2.5×10⁻⁵)/(-1.75×10⁻⁶)
  • E = -14.3 N/C downward

Since the charge is negative, the field and the force are in opposite directions.

Hence, the magnitude and direction of the electric field is  -14.3 N/C Downward.

Learn more about Electric Field here: https://brainly.com/question/11509296

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