A charge of 28.0 nC is placed in a uniform electric field...

A charge of 28.0 $\mathrm{nC}$ is placed in a uniform electric field that is directed vertically upward and has a magnitude of $4.00 \times 10^{4} \mathrm{V} / \mathrm{m}$ . What work is done by the electric force when the charge moves (a) 0.450 $\mathrm{m}$ to the right; (b) 0.670 $\mathrm{m}$ upward; (c) 2.60 $\mathrm{m}$ at an angle of $45.0^{\circ}$ downward from the horizontal?

Key Concepts

Work in Electric Fields

This concept involves calculating the work done on a charged particle within an electric field. The work is determined by the product of the charge, the magnitude of the electric field, and the displacement component in the direction of the field. This is mathematically expressed using the dot product of the electric field vector and the displacement vector.

Dot Product in Work Calculations

The dot product is a mathematical operation used to project one vector onto another. In the context of work done by an electric force, it identifies the portion of the displacement that aligns with the electric field direction, which is crucial because only this component contributes to the work done.

Uniform Electric Field

A uniform electric field is characterized by having the same magnitude and direction throughout the region. Such a field simplifies the analysis because the force acting on a charge remains constant regardless of the charge's position, allowing for straightforward work calculations using the formula work = qE·d.

Vector Projection

Vector projection involves decomposing a displacement vector into components along specific directions. For work calculations in electric fields, it is important to extract the displacement component parallel to the field vector, since this determines the actual work performed by the field on the moving charge.

Influence of Displacement Direction

The direction of displacement relative to the direction of the electric field is critical in determining the work done. If the displacement is perpendicular to the electric field, no work is done; if it is parallel, the full force contributes to the work; and for any oblique angle, only the parallel component affects the work.

Transcript

00:01 In this problem, we have a constant electric field that we're told is equal to 4 times 10 to the 4th volts per meter.

00:16 Okay? and we have a point charge in that electric field.

00:22 I'll just put it there for now.

00:24 And that point charge is 28 nanoculoms, so 28 times 10 to the negative 9 cool.

00:34 Now i'm going to choose a coordinate system because we're going to need it later on and i'm just going to pick the traditional coordinate system here.

00:43 Okay, and now we're asked to calculate the work for different displacements of this charge.

00:51 So if you remember work is equal to the integral of force.

00:58 D .l and since we have an electric field here, that force is is just replaced by q times the electric field dotted with dl.

01:12 Okay? so now our charge is a constant.

01:16 Our electric field is a constant.

01:18 So we can pull out both of those constants and do this integral.

01:22 So this is simply equal to, and as well i'm gonna plug in what the dot product is equal to.

01:28 So qe, and then with that dot product, we get a cosine of the angle between the electric, field and the displacement vector times the integral of dl, which is just the length that that vector moves.

01:45 So i'm just going to call it l, since we have dl here already.

01:50 So now, in each case, we just need to calculate this.

01:54 So for part a, we're told that the particle moves 0 .45 meters to the right.

02:03 So this would be l, l vector is equal to 0 .45 meters x -hat.

02:15 Okay.

02:17 So now we can plug this in, work a is equal to qe times 0 .45 times cosine of the angle...

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