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An electric field of magnitude S.25 105 N/C points due south at a certain location. Find the magnitude and direction of the force on a-6 charge at this location.

Respuesta :

Answer:

F=3.15 N pointing south

Explanation:

For the magnitude:

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

F = qE = 6*10x^{-6} *5.25*10^{5}[/tex]

[tex]F=3.15 N[/tex]

For the direction:

In the equation F=qE, both F and E are vectors while q is the module of the charge, this means that the direction of the electric field is the one that gives the direction of the force, in this case both point south

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