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 $\vec{F}_{d}$ on the 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 $F_{d^{\prime}} / F_{g}$ in this case?

Key Concepts

Force Ratio Analysis

Analyzing the ratio of the electrostatic force on a charged particle to its gravitational force is key in understanding which force dominates the particle’s behavior in a field. This comparison, often represented as F_e/F_g, enables the assessment of whether electric or gravitational effects are more significant in a given context, which is critical in fields ranging from atmospheric electricity to particle physics.

Sign of Charge and Force Direction

The sign of the electrical charge plays a crucial role in determining the direction of the electrostatic force exerted by an electric field. For a positive charge, the force is in the same direction as the electric field, whereas for a negative charge, the force is opposite the field’s direction. This distinction is essential for correctly analyzing the motion and behavior of different charged particles in an electric field.

Gravitational Force

Gravitational force is the attractive force that acts on any mass due to gravity, calculated using the equation F_g = mg, where m is the mass of the object and g is the acceleration due to gravity. In many physical situations, especially at the Earth’s surface, this force provides a baseline for comparing and analyzing other forces acting on an object, such as the electrostatic force.

Electric Field

The electric field is a vector field that describes the force per unit charge exerted at a given point in space. It is defined by the relationship E = F/q, meaning that its magnitude is determined by dividing the magnitude of the force by the magnitude of the charge producing it. This concept is fundamental in linking electric forces with the properties of the mediums in which they exist.

Electrostatic Force on a Charge

The electrostatic force experienced by a charged particle in an electric field follows the equation F = qE, where q is the charge of the particle and E is the electric field. This equation underlines how the magnitude and direction of the force depend not only on the electric field strength but also on the magnitude and sign of the charge. A positive charge experiences a force in the direction of the field while a negative charge sees it in the opposite direction.

Transcript

00:01 So well in this question we have given the magnitude of the charges over here and and force acting on that charges is given so f equals to what 3 into 10 to the power of minus 6 newton and the charge q equals to minus of 2 into 10 to the power of minus 9 column so we need to find the forces so what we can do over here is that according to the convention the test charge is positive by negative obviously in the strength of electric field at some point.

00:32 So if the force is given, if force is given, and obviously we can calculate that electric field, how, how, if the force is according to convention the test charges, what is the assistance of electric field at some point is equal to the force point in the first point if the charge is negative, electrostatic force will be negative and because it act in the opposite direction of the electric field.

00:58 So since, we know already that this one so we can find the strength in the first question a strength of the electric field by you know that what f equals to qe so e equals to f by q or if obviously we need to just find the direction so minus q implies just the the direction of force which will react if electric field in this direction then the force will be in this direction and electric field will in this direction so we can just put the magnitude over here so force 3 into 10 minus 6 upon our minus and over here so minus minus minus get plus 2 into 20 or minus 9 so it comes out to be 1 .5 into 10 to the power of 3 newton per column okay now coming back to the next part now we have to find the force and force experiences on the proton i think so um electric field force experiences on the proton so we know that f equals to nothing but 2 into e.

02:00 So charge on proton is 1 .6 into 10 to the power of minus 19 and into electric field.

02:05 Electric field is 1 .5 to the power of 3.

02:08 So it comes out to be, i think, 2 .6 into 10 to the power of minus 16 newton.

02:16 Okay.

02:17 Now coming back to the next part, what we have to find in the next part is.

02:30 So in the second part, we have to calculate magnitude in the direction of electrostatic force.

02:35 So in the c part, we need to find the direction...

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