Physics, 09.07.2019 03:30 itsgiovanna

A0.15g plastic bead is charged by the addition of 1.0×1010 excess electrons, what electric field e⃗ (strength) will cause the bead to hang suspended in the air? what electric field e⃗ (direction) will cause the bead to hang suspended in the air, up or down?

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Answer from: hendo617

The electric field will cause the bead to hang suspended in the air is in downward direction with magnitude,

$E=9.186\times10^{5}\rm V/m$

What is electric field?

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge. It can also be given as,

q=ne

Here (e) is the charge on an electron.

The mass of the plastic bead is 0.15 gram.

The charge added to the plastic bead is $1.0\times10^{10} \rm C$ .

As, the charge on one electron is equal to the $1.6\times10^{-19}\rm C$ . Thus, put these values in the above formula as,

$q=1.0\times10^{10}(1.6\times10^{-19})\\q=(1.6\times10^{-9})\rm C\\$

The electric filed for the balance is the ratio of the object and the charge it is experienced on it.

The electric field E (strength) will cause the bead to hang suspended in the air can be given as,

$E=\dfrac{0.15\times10^{-3}\times9.8}{1.6\times10^{-9}}\\E=9.186\times10^{5}\rm V/m$

As the charge on the bead is negative in the direction, thus the electric field strength should be applied in opposite direction, which is downward direction.

Hence, the electric field will cause the bead to hang suspended in the air is in downward direction with magnitude,

$E=9.186\times10^{5}\rm V/m$

Learn more about electric field here;

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A0.15g plastic bead is charged by the addition of 1.0×1010 excess electrons, what electric field e⃗

Answer from: ameliaparry123p2eupk

The magnitude of the electric field for the bead to hang suspended in air is 9.19×10⁵V/m and the field is directed downwards.

If n electrons each of charge e are added to the bead, the total charge q on the bead is given by

q=ne

Substitute 1.0×10¹⁰ for n and 1.6×10⁻¹⁹C for e.

$q=ne\\ =(1.0*10^1^0)(1.6*10^-^1^9C)\\ =1.6*10^-^9C$

The bead ha a negative charge of magnitude 1.6×10⁻⁹C.

The bead has a mass m and its weight mg is directed downwards. The bead can be balanced in an electric field of strength E if the force it experiences due to the electric field is directed upwards and is equal to the magnitude of its weight. This is shown in the diagram attached.

For equilibrium,

mg=qE

Rewrite the expression for E.

$E=\frac{mg}{q}$

Substitute 0.15 g = 0.15×10⁻³kg for m, 9.8 m/s² for g and 1.6×10⁻⁹C for q.

$E=\frac{mg}{q}\\ =\frac{(0.15*10^-^3kg)(9.8m/s^2)}{(1.6*10^-^9C)} \\ =9.1875*10^5V/m$

The magnitude of the electric field required to balance the bead is 9.19×10⁵V/m.

The bead is negatively charged. For the bead to experience a force in the upward direction, the electric field is to be directed down wards.

The electric field is directed downwards.

Answer from: Quest

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Answer from: Quest

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