A charged cloud system produces an electric field in the air...

A charged cloud system produces an electric field in the air near Earth's surface. A particle of charge $-2.0 \times 10^{-9} \mathrm{C}$ is acted on by a downward electrostatic force of $3.0 \times 10^{-6} \mathrm{~N}$ when placed in this field. (a) What is the magnitude of the electric field? What are the (b) magnitude and (c) direction of the electrostatic force $\vec{F}_{e l}$ on a proton placed in this field? (d) What is the magnitude of the gravitational force $\vec{F}_{g}$ on the proton? (e) What is the ratio $F_{e l} / F_{g}$ in this case?

Key Concepts

Charge Polarity and Force Direction

Charge polarity plays a crucial role in determining the direction of the force experienced by a particle in an electric field. A positively charged particle will experience a force in the direction of the electric field, while a negatively charged particle will experience a force in the opposite direction. Understanding this concept is vital for correctly interpreting the behavior and motion of charged particles under various electric field conditions.

Force Ratio

The ratio of the electrostatic force to the gravitational force provides insight into the relative strengths of these forces on a charged particle. By comparing these forces, one can assess how dominant the electrical effects are compared to gravitational effects for given charges and masses. This concept is especially important in fields like atomic physics and astrophysics, where it is often used to justify when certain forces can be neglected in favor of others.

Electric Field

The electric field is defined as the force experienced per unit charge. In problems involving charges and forces, the magnitude of the electric field can be calculated by dividing the force acting on a test charge by the magnitude of that charge, regardless of its sign. This concept is fundamental in understanding how charged objects influence their surroundings and how other charged particles will behave in the field.

Electrostatic Force

The electrostatic force on a charge placed in an electric field is determined by the equation F = qE, where q is the charge and E is the electric field. This relationship shows that the force acting on a particle is directly proportional to both its charge and the strength of the electric field it is in. This concept is essential for analyzing the interactions between charged particles in uniform or non-uniform fields.

Gravitational Force

Gravitational force is the attractive force that acts between masses, calculated using the equation F = mg, where m is the mass of an object and g is the acceleration due to gravity. Although typically much weaker than the electrostatic force at the atomic and subatomic scales, gravitational force is significant in contexts where particle mass is considered, especially when comparing different forces acting on the same particle.

Transcript

0:00 Hi there.

00:01 So for this problem, we have a particle that has a charge of minus 2 times 10 to the minus 9 columns.

00:17 And it's acted on by a downward electrostatic force.

00:23 So we are given electrostatic force, which is equal to 3.

00:31 Times 10 to the minus 6 neutons when it's placed in that field.

00:43 Now what we need to determine is the magnitude of the electric field.

00:53 So for the first part, we need the magnitude of the electric field.

00:58 Now, we know that the electric field is defined as the force overt, the charge.

01:11 Now in here we just need to simply substitute those values.

01:16 Of course this is the magnitude of the charge so we will have that this is 3 times 10 to the minus 6 neutons over 2 times 10 to the minus 9 column.

01:30 So from here we obtain an electric field of 1 .5 times 10 to the 3 newtons per column so this is the solution for this part of the problem now you can also put this in a vector notation and since the force points downward and the charge is negative the electric field must point upward upward now for part b of this problem about the magnitude of the electrostatic force on a proton that is placed in this field.

02:20 So we need to find that for a proton.

02:26 So again, we know that the force, the electric field force, is defined as the charge times the electric field.

02:36 Now we have calculated the electric field, but in this case, the charge, corresponds to the charge of a proton.

02:44 And we know that the charge of a proton is simply the charge of an electron but positive.

02:53 So we will have 1 .6 times 10 to the minus 19 columns.

02:59 That's the charge of a proton times the electric field that we obtain.

03:07 It is 1 .5 times 10 to the 3 newton's per column...