Question: An alpha particle (the nucleus of a helium atom) has a mass of $\mathbf{6}\mathbf{.}\mathbf{64}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}$and a charge of +2e. What are the (a) magnitude and (b) direction of the electric field that will balance the gravitational force on the particle?

Mass of the alpha particle,$m=6.64\times {10}^{-27}kg$

Charge of the alpha particle,$q=+2e$

The force on the object due to gravity, also called as weight of the object, is equal to the product of mass and the gravitational acceleration. When the charged particle is placed in an electric field, the electrostatic force acting on the particle is equal to the product of charge and electric field.

Using the concept of force due to gravity on the alpha particle and relating the value to the electric field produced by the particle, we can get the magnitude of the electric field with its direction.

Formulae:

The force due to gravity, F = mg (i)

The force and electric field relation, F = qE (ii)

Vertical equilibrium of forces leads to the balance of the forces due to gravity and due to electric filed. Thus, using equation (i) in equation (ii), we can get the magnitude of the electric field as follows:

$$\begin{gathered} \left|\underset{E}{\to }\right|=\frac{mg}{2e} \\ =\frac{\left(6.64\times {10}^{-27}kg\right)\left(9.8m/s^2\right)}{2(1.60\times {10}^{-19}C)} \\ =2.03\times {10}^{-7}N/C \end{gathered}$$

Hence, the value of the electric field is$2.03\times {10}^{-7}N/C$

Since the force of gravity is downward, then the force due to the electric field must point upward. Since q > 0 in this situation, this implies the electric field must itself point upward.