A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 StartFraction N over C EndFraction. What is the electric force acting on the charge?

Respuesta :

Answer:

0.9 N

Explanation:

The electric force, F, is the product of the electric field strength, E, and the charge i.e.

[tex]F = E\cdot Q[/tex]

From the question,

[tex]E = 2.0\times 10^4\text{ N/C}[/tex]

[tex]Q = 4.5\times10^{-5}\text{ C}[/tex]

[tex]F = (2.0\times10^4\text{ N/C})(4.5\times10^{-5}\text{ C}) = 9.0\times10^{-1}\text{ N} = 0.9\text{ N}[/tex]

Answer: 0.9

Explanation: