Solving Electric Field Problems

The oil drop has a charge of -1.65 X 10 ‾17 C. It has 3.3 X 10 ‾19 extra electrons.In summary, The oil drop has a charge of -1.65 X 10 ‾17 C and contains 3.3 X 10 ‾19 extra electrons. The drop remains stationary due to the forces of gravity and the electric field canceling each other out.
  • #1
rhaylor
1
0
I'm just getting started here and need a little help. Need a general method to figure problems like these:

If a drop of mass 1.1 X 10 ‾14 kg remains stationary in an electric field of 1.68 X 10 N/C:

(a) What is the charge of this oil drop?

(b) How many extra electrons are on this particular oil drop (given the presently-known charge of the electron)?

Any help is deeply appreciated.

rhaylor :redface:
 
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  • #2
rhaylor said:
If a drop of mass 1.1 X 10 ‾14 kg remains stationary in an electric field of 1.68 X 10 N/C:


Well... What forces act on that oil drop? It is here on the Earth and it has mass, so it can not avoid gravity. And it is in an electric field, and it has charge, although unknown yet, but is does not move so the forces must cancel each other...

ehild
 
  • #3


Hi rhaylor,

Solving electric field problems can seem daunting at first, but with a general method, it becomes much more manageable. Here are some steps you can follow:

1. Identify the known values and unknown value in the problem. In this case, we know the mass of the drop (1.1 x 10^-14 kg) and the electric field (1.68 x 10^3 N/C), and we need to find the charge and the number of electrons.

2. Recall the equation for electric force, which is F = qE, where q is the charge and E is the electric field. We can rearrange this equation to solve for q: q = F/E.

3. Plug in the known values into the equation and solve for q. In this case, q = (1.1 x 10^-14 kg)(1.68 x 10^3 N/C) = 1.848 x 10^-17 C.

4. To find the number of electrons, we can use the fact that one electron has a charge of 1.6 x 10^-19 C. So, the number of electrons on the oil drop would be 1.848 x 10^-17 C / 1.6 x 10^-19 C = 115.5 electrons.

I hope this helps! Remember to always identify the known and unknown values, use the relevant equations, and double check your units to make sure they cancel out correctly. Best of luck with your studies!
 

1. What is an electric field?

An electric field is a physical field that surrounds charged particles and exerts a force on other charged particles within its range. It is represented by arrows pointing in the direction of the force that would be exerted on a positive test charge placed in the field.

2. How do you calculate the electric field at a point?

The electric field at a point can be calculated by dividing the force exerted on a test charge by the magnitude of the test charge. Mathematically, it can be represented as E = F/q, where E is the electric field, F is the force, and q is the test charge.

3. What is the unit of electric field?

The unit of electric field is newtons per coulomb (N/C) in SI units. In other systems, it can also be represented as volts per meter (V/m) or pascals per meter (Pa/m).

4. How can you determine the direction of an electric field?

The direction of an electric field is determined by the direction of the force that would be exerted on a positive test charge placed in the field. The direction of the electric field is always in the same direction as the force on a positive test charge and opposite to the force on a negative test charge.

5. What are some common applications of solving electric field problems?

Solving electric field problems is an important tool in understanding and designing electrical systems. It is used in a wide range of applications such as designing electronic circuits, calculating the forces in charged particle accelerators, and understanding the behavior of lightning strikes.

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