Two point charges are located on a coordinate system as...

Two point charges are located on a coordinate system as follows: $Q_{1}=-4.5 \mu \mathrm{C}$ at $x=1.00 \mathrm{cm}$ and $y=1.00 \mathrm{cm}$ and $Q_{2}=6.0 \mu \mathrm{C}$ at $x=3.00 \mathrm{cm}$ and $y=1.00 \mathrm{cm}$ (a) What is the electric field at point $P$ located at $x=$ $1.00 \mathrm{cm}$ and $y=4.00 \mathrm{cm} ?$ (b) When a tiny $5.0 \mathrm{g}$ particle with a charge of $-2.0 \mu \mathrm{C}$ is placed at point $P$ and released, what is its initial acceleration?

Two point charges are located on a coordinate system as follows: $Q_{1}=-4.5 \mu \mathrm{C}$ at $x=1.00 \mathrm{cm}$ and $y=1.00 \mathrm{cm}$
and $Q_{2}=6.0 \mu \mathrm{C}$ at $x=3.00 \mathrm{cm}$ and $y=1.00 \mathrm{cm}$
(a) What is the electric field at point $P$ located at $x=$ $1.00 \mathrm{cm}$ and $y=4.00 \mathrm{cm} ?$ (b) When a tiny $5.0 \mathrm{g}$ particle with a charge of $-2.0 \mu \mathrm{C}$ is placed at point $P$ and released, what is its initial acceleration?

Key Concepts

Vector Addition and Decomposition

Because electric fields are vector quantities, it is necessary to decompose them into their horizontal and vertical components to accurately calculate the net field. This concept involves breaking down vectors into components, summing the contributions in each direction, and then recombining them to get the resultant vector.

Electric Field of a Point Charge

This concept involves understanding that a point charge generates an electric field in the surrounding space, which is described by Coulomb’s law as E = kQ/r². The field points radially outward from a positive charge and inward toward a negative charge, and its magnitude decreases with the square of the distance from the charge.

Superposition Principle

The superposition principle states that when multiple charges are present, the total electric field at any given point is the vector sum of the electric fields produced by each charge independently. This is essential for calculating the net electric field at a point due to several point charges.

Coulomb’s Law

Coulomb's law provides the basis for calculating the electric field and force between two point charges. It quantifies the interaction by stating that the force is directly proportional to the product of the charges and inversely proportional to the square of the separation distance, which underpins the calculation of the electric field from a point charge.

Force on a Charged Particle

The force experienced by a charged particle in an electric field is given by F = qE, where q is the charge of the particle and E is the electric field. This concept links the electric field to the mechanical effect, representing how the particle will move under the influence of the field.

Newton's Second Law of Motion

Newton's second law, F = ma, connects the force exerted on an object to its acceleration and mass. In problems involving forces from electric fields, it allows us to determine the acceleration of a charged particle once the force acting on it is known.

Transcript

00:01 In the given problem, first of all we draw y -axis x -axis.

00:13 Now in this system, in this x -y plane, there is a charge q1 having a value of minus 4 .50 microculum put at a position having coordinates 1 .00 cm, comma, and 1 .00 centimeter.

00:40 So that point is supposed to be placed here, present here.

00:47 And we assume this point to be a, at which we put this charge q1.

00:54 Now there is another charge q2 which is equal to 6 .00 micropulum and it is put at a location 3 .00 cm means 3 .00 along x -axis and 1 .00 centimeter means that will be along y -axis so its position along y -axis is the same so it will remain here but along x -axis if this distance is 1 .0 centimeter then this distance is 3 centimeter from means the gap between a and p is actually 2 .0 centimeter and at this point b we have put the charge q 2 now there is a point p which is at 1 .00 centimeter comma 4 .0 centimeter means the distance along x -axis is just 1 and distance along y -axis is 3 sorry 4 so this is point p this and this ap then will become equal to 4 minus 1 means this is 3 .0 centimeter while this was 2 .0 centimeter now in the first part of the problem we have to find electric field the total electric field at point p.

02:36 So actually there will be two electric fields at point p.

02:39 One because of this positive charge put at a here this was sorry this was negative charge and this was positive charge.

02:48 So due to this negative charge put at a electric field will be towards the negative charge.

02:55 This is ea and electric field due to positive charge it goes away.

03:01 It means it will be like this e b.

03:05 To find the net electric field we will need the angle between them also.

03:10 And to find this angle, first of all, we find this angle and let this angle be alpha because here this pab will become a right angled triangle...

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