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

Respuesta :

Force experienced by a charge in an electric field is given by,

[tex]F_{electric}=qE[/tex]

where q is the charge, E is the electric field,

Given,

[tex]\begin{gathered} q=5.0\times10^{-4}C \\ F_{electric}=2.5\times10^{-4}N \end{gathered}[/tex]

Hence,

[tex]\begin{gathered} 2.5\times10^{-4}=5.0\times10^{-4}\times E \\ E=0.5\text{ N/C} \end{gathered}[/tex]

Result: 0.5 N/C

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