A uniform electric field has magnitude 240 N / C and is directed to the right. A particle with c

A uniform electric field has magnitude 240 N / C and is...

A uniform electric field has magnitude $240 \mathrm{N} / \mathrm{C}$ and is directed to the right. A particle with charge $+4.2 \mathrm{nC}$ moves in this field. For the given motion of the particle, find (a) the electric force that acts on the particle; (b) the potential difference through which the particle moves; (c) the change in the particle's potential energy; and (d) the work done on the particle by the electric field. The particle moves along a straight line from $a$ to $b$.

A uniform electric field has magnitude $240 \mathrm{N} / \mathrm{C}$ and is directed to the right. A particle with charge $+4.2 \mathrm{nC}$ moves in this field. For the given motion of the particle, find
(a) the electric force that acts on the particle;
(b) the potential difference through which the particle moves;
(c) the change in the particle's potential energy; and
(d) the work done on the particle by the electric field.
The particle moves along a straight line from $a$ to $b$.

Key Concepts

Work Done by the Electric Field

Work done by the electric field on a charged particle is the energy transferred as the particle moves within the field. It is equal to the change in the particle's kinetic energy and is calculated by the product of the force acting on the charge and the displacement in the direction of the force, reflecting the interplay between force, displacement, and energy transfer.

Electric Potential Energy

Electric potential energy in the context of a charged particle within an electric field represents the energy stored due to the position of the charge in the field. Changes in this potential energy can be understood by the relationship ?U = q?V, linking the work done on the particle and the resulting energy change.

Electric Potential Difference

Electric potential difference, often simply called voltage, is the change in electric potential energy per unit charge between two points in an electric field. It quantifies the work done by or against the field when moving a charge between those points and is directly related to the product of the electric field strength and the displacement in the direction of the field.

Electric Force on a Charge

The electric force acting on a charged particle in an electric field is determined by the equation F = qE, where q is the charge of the particle and E is the electric field strength. This concept helps in understanding how a charged particle responds to the field, with the force's direction depending on the sign of the charge in relation to the field direction.

Uniform Electric Field

A uniform electric field is a region of space where the electric field vector has constant magnitude and direction at every point. This means that the force exerted on a charge placed in the field is the same in magnitude and direction regardless of its location, making the analysis of motion and energy changes straightforward using simple linear equations.

Transcript

00:01 High in the given problem there is a uniform electric field e in vector form and it is having a magnitude of 240 newton per coulain.

00:22 In this electric field there are two points a and b having a distance are between them which is having a magnitude of 0 .25 meter.

00:42 Now a charge q is equal to plus 4 .2 nanoculum is shifted from a to b.

01:00 So in this process in the first part of the problem we have to find the electrostatic force experienced by this chart particle as we know elective field is always from positive to negative so this positive charge which is moving from a to b suppose this is the positive charge which is being moved from a to b this charge will be experiencing a force of repulsion from a to b means positive will be repelling it away towards negative so the force experienced by this charge will be given by the product of charge with the electric field so it will be given by 4 .2 into 10 dash per minus 9 coularm multiplied by distance which is 0 point sorry multiplied by electric field which is 240 newton per coularm and here this was coularm so canceling this coulin, finally this electrostatic force experienced by the charged particle will come out to be 1 .008 into 10 dash to the power minus 6 newton or we can say this is f is equal to 1 .0 micron.

02:27 Answer for the first part of the problem and this force is being experienced by the charge particle towards right.

02:40 Now in the second part of the problem, we have to find the potential difference between these two points.

02:49 So that potential difference is given by the relation between potential difference and electric field which says potential difference is equal to the product of electric field with the distance between the points.

03:02 So in magnitude it will be given by 200.

03:06 40 newton perpula multiplied by the distance which is 0 .25 meter...

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