An object having a net charge of 24µC is placed in a uniform electric ﬁeld of 610 N/C directed vertically. What is the mass of this object if it “ﬂoats” in the ﬁeld?

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An object having a net charge of 24µC is placed in a uniform electric ﬁeld of 610 N/C directed vertically. What is the mass of this object if it “ﬂoats” in the ﬁeld?

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Answer 1 · 2016-02-20T04:30:11

The mass of the object is $1.5 \times 10^{- 3} k g = 1 .5 g .$ $1.5×10^-3 kg = 1 .5g.$
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Explanation:

The forces acting on the mass are shown in image that I uploaded. The force of gravity points downward and has magnitude mg (m is the mass of the object) and the electrical force acting on the mass has magnitude $F = | q | E$ $F = |q|E$ , where $q$ $q$ is the charge of the object and $E$ $E$ is the magnitude of the electric ﬁeld. The object “ﬂoats”, so the net force is zero. This gives us: $| q | E = m g$ $|q|E = mg$ Solve for mass,

$m = \frac{| q | E}{g} = (24 \times 10^{- 6} C) \frac{610 \frac{k g \cdot \frac{m}{s^{2}}}{C}}{9.80 \frac{m}{s^{2}}} = 1 .5 \times 10 - 3 k g$ $m = (|q|E) /g = (24×10^-6 C)(610 (kg*m/s^2)/C)/ (9.80 m/s^2) =1 .5×10−3 kg$

The mass of the object is $1.5 \times 10^{- 3} k g = 1 .5 g .$ $1.5×10^-3 kg = 1 .5g.$

enter image source here

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An object having a net charge of 24µC is placed in a uniform electric ﬁeld of 610 N/C directed vertically. What is the mass of this object if it “ﬂoats” in the ﬁeld?

1 Answer

Explanation:

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