A speck of charged dust with a mass of 39mu g. is kept levitating by putting it in an electric field. If you must apply a field of 10 N/C to keep it suspended in mid-air, what is the charge on the speck of dust?
Can someone please help soon? I got this wrong on my last test because I'm terrible at physics and I'm trying not to make the same mistake on the exam. An explanation would be great.
To find the charge on the speck of dust, we can use the formula:
F = q * E
where F is the force applied on the speck of dust, q is the charge on the dust particle, and E is the electric field.
We are given that the electric field E = 10 N/C, and we want to find the charge q.
The force F is the weight of the dust particle, which can be calculated using the formula:
F = m * g
where m is the mass of the speck of dust and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the mass of the dust particle is 39 μg (39 * 10^-6 g), we can convert it to kg:
m = 39 * 10^-9 kg
Now we can find the force F:
F = m * g
F = (39 * 10^-9) * 9.8 N
F = 3.822 * 10^-7 N
We know that the force applied on the speck of dust is equal to the force due to the electric field:
F = q * E
Substituting the values we know:
3.822 * 10^-7 N = q * 10 N/C
Now, let's solve for q:
q = (3.822 * 10^-7 N) / (10 N/C)
q = 3.822 * 10^-8 C
Therefore, the charge on the speck of dust is 3.822 * 10^-8 C.
To avoid making mistakes in physics, it is important to understand and apply the relevant formulas, units, and conversion factors. Practice solving similar problems and review the concepts regularly. You can also seek additional resources like textbooks, online tutorials, or engaging in study groups to enhance your understanding of physics.