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A point charge has q=1.0×10-6
coul. Consider point A 2.0 meters distant and
point B 1.0 meter distant from the charge.
ii) If point A is on east side of the charge and point B on north side of the
charge, what is the potential difference VA-VB? Draw a diagram of the
arrangement.

Respuesta :

IsuruYahampath

Since the 2 points form a triangle with hypothenuse of √5 [ √(1²+2²)],

I guess apply the formula :

[tex]v = kq \div r[/tex]

with r as √5 and q as 1x10^-6

not sure about this answer tho

draw a diagram first to understand better

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