Name: Dipole in an Electric Field
Uploaded: 2019-07-19
Duration: 11 min 59 s
Description: Dipoles exists everywhere. They apply to both electricity and magnetism in physics. Here in this video, let us explore how an electric dipole interacts with...

Question

An alpha particle (the nucleus of a helium atom) has a mass of

6.64 \times 10^{- 27} kg

and a charge of

+ 2 e .

What are the

(a)

magnitude and

(b)

direction of the electric field that will balance the gravitational force on the particle?

(c)

If the field magnitude is then doubled, what is the magnitude of the particle's acceleration?

Accepted Answer

Alpha particle with mass of $m = 6.64 \times 10^{- 27} kg$ and with a charge of $q = + 2 e$ , we need to find the electric field in order to balance the gravitational force.

(a) To balance the particle, the force sum must be equal zero, this lead is to,

$F_{g} = F_{E}$ ,

where, $F_{E}$ is the given by $q E$ , and $F_{g}$ is the gravitational force and it is given by $m g$ , substitute we get,

$m g = q E$

$\Rightarrow m g = 2 e E$ .

Rewrite for $E$ we get,

$E = \frac{m g}{2 e}$ .

Substitute with the given values are get,

$E = \frac{(6 .64 \times 10^{- 27}) (9 .8)}{2 (1 .6 \times 10^{- 19})}$

$\Rightarrow E = 2 .0335 \times 10^{- 7} N C^{- 1}$ .

(b) Since the charge is positive, the electric force has the same direction of the electric field according to $\vec{F} = q \vec{E}$ , this the direction of electric field is upward, must itself point upward.

(c) If the magnitude of electric field is doubled, it will also double the electric force which will be equal to $2 m g$ in magnitude. So particle wil now accelerate with acceleration $g$ upwards.

Important Questions on Electric Fields

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