What upward pointing ⃗Efield magnitude would be required to levitate a bare carbon nucleus (assume carbon-12 with no electrons, just the nucleus) near the surface of the earth in N/C? For this problem, it is useful to draw a free body charge of the Carbon nucleus take into account the gravitational force and the electric force acting on the carbon nucleus
To find the upward pointing electric field magnitude required to levitate the bare carbon nucleus, we need to balance the gravitational force acting downwards with the electric force acting upwards.
1. First, let's consider the gravitational force acting on the carbon nucleus. The gravitational force is given by the equation:
F_gravity = m * g
where F_gravity is the gravitational force, m is the mass of the carbon nucleus, and g is the acceleration due to gravity, approximately 9.8 m/s².
The mass of the carbon-12 nucleus (C-12) is approximately 1.993 x 10⁻²⁶ kg.
Therefore, the gravitational force on the carbon nucleus is:
F_gravity = (1.993 x 10⁻²⁶ kg) * (9.8 m/s²)
2. Next, let's consider the electric force acting on the carbon nucleus. The electric force is given by the equation:
F_electric = q * E
where F_electric is the electric force, q is the charge of the carbon nucleus, and E is the electric field magnitude.
Since the carbon nucleus has no electrons, it has a charge equal to the elementary charge e = 1.602 x 10⁻¹⁹ C.
Therefore, the electric force on the carbon nucleus is:
F_electric = (1.602 x 10⁻¹⁹ C) * E
3. For levitation to occur, the electric force must balance the gravitational force. Thus, we have:
F_electric = F_gravity
(1.602 x 10⁻¹⁹ C) * E = (1.993 x 10⁻²⁶ kg) * (9.8 m/s²)
4. Now, we can solve for the electric field magnitude:
E = [(1.993 x 10⁻²⁶ kg) * (9.8 m/s²)] / (1.602 x 10⁻¹⁹ C)
Calculate the numerical value to find the upward pointing electric field magnitude required to levitate the bare carbon nucleus near the surface of the earth.