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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?

Short Answer

Expert verified
The work done by the electric force when the charge moves (a) 0.450 m to the right is 0 J; (b) 0.670 m upward is 7.536 x 10^-4 J; and (c) 2.60 m at an angle of 45.0 degrees downward from the horizontal is 2.576 x 10^-3 J.

Step by step solution

01

Initialize given parameters

First, identify the given parameters. The charge \(q = 28.0 \, nC = 28.0 \times 10^{-9} \, C\) (converted to Coulombs) and the electric field \(E = 4.00 \times 10^{4} \, V/m\). The distances and their directions are given as: \(d_1 = 0.450 \, m\) (right = perpendicular to E), \(d_2 = 0.670 \, m\) (upward = along E), and \(d_3 = 2.60 \, m\) (45 degrees downward = angle with E).
02

Case (a) - Calculate the work done by electric force on charge displaced rightward

The work done is calculated by using formula \( \text{W} = qEd\cos(\Theta)\). Because the displacement is rightward, that is perpendicular to the electric field (\(E\)), the angle \(\Theta\) between them is 90 degrees. Substituting into the equation gives \( \text{W} = 28.0 \times 10^{-9} \times 4.00 \times 10^{4} \times 0.450 \times \cos(90^{\circ}) = 0 \, J\), because the cosine of 90 degrees is zero.
03

Case (b) - Calculate the work done by electric force on charge displaced upward

Again using the formula \( \text{W} = qEd\cos(\Theta)\) and noting that the displacement is upward and along the direction of the electric field, the angle \(\Theta\) between them is 0 degrees. So, \( \text{W} = 28.0 \times 10^{-9} \times 4.00 \times 10^{4} \times 0.670 \times \cos(0^{\circ}) = 7.536 \times 10^{-4} \, J\).
04

Case (c) - Calculate the work done by electric force on charge displaced at 45.0 degrees downward from the horizontal

Once again using the formula \( \text{W} = qEd\cos(\Theta)\), now the displacement \(d3\) at 45 degrees forms an angle \(\Theta\) of 45 degrees with the electric field. So, \( \text{W} = 28.0 \times 10^{-9} \times 4.00 \times 10^{4} \times 2.60 \times \cos(45^{\circ}) = 2.576 \times 10^{-3} \, J\). Note that the cosine of 45 degrees is \( \frac{1}{\sqrt{2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force
Imagine you're holding two magnets near each other. You can feel a pull or a push even when they are not touching—that's a force at a distance. Similarly, electric force works between charged particles. Defined by Coulomb's law, this force can attract or repel charged particles, such as electrons or protons, without physical contact.

Mathematically, Coulomb's law is expressed as \( F = k \frac{{|q_1 q_2|}}{{r^2}} \), where \( F \) is the electric force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the electric charges, and \( r \) is the distance between the centers of the two charges. Understanding electric force is key to work calculation, as it helps explain how charges interact within an electric field.
Work Done by Electric Force
When we talk about work in physics, we’re referring to the energy transferred to or from an object via the force causing the object to move. Work done by electric force, then, is the energy transferred by the electric force that moves a charge over a distance.

Think of it like pushing a shopping cart: the work you do depends on the force you apply and the distance you push it. For electric force, this is calculated using the formula \( W = qEd\cos(\Theta) \), where \( W \) is the work done, \( q \) is the charge moving through the electric field with magnitude \( E \), \( d \) is the distance the charge moves, and \( \Theta \) is the angle between the direction of the force and the movement. The cosine factor becomes critical as it adjusts the work done based on the charge's path in relation to the electric field direction.

In cases where the charge moves perpendicular to the field, like in the provided exercise part (a), no work is done by the field, as the force and movement directions are 90 degrees apart, making \( \cos(90^\circ) \) equals zero. In part (b), the work is maximal due to the alignment of movement and force, making \( \Theta = 0^\circ \) and thus \( \cos(0^\circ) = 1 \). This relationship signifies not just a quantity of energy transfer, but its directionality as well.
Uniform Electric Field
Just as a calm ocean allows for smooth sailing, a uniform electric field provides a consistent force on a charged particle placed within it. A uniform electric field has the same magnitude and direction throughout its extent. Visually, you could represent it as equally spaced, parallel lines, never crossing—much like the lines on a piece of graph paper.

Mathematically, the force \( F \) experienced by a charge \( q \) in a uniform electric field \( E \) is given by \( F = qE \). There's no need to worry about where in the field the charge is located or the path it takes—all that really matters is the field's strength and the charge's magnitude.

In practical terms, to calculate the force on a charge in such a field, you only need the charge value and the field strength. The uniformity simplifies calculations, especially in cases like the provided exercise where work depends on direction—upward, downward, or perpendicular. It allows us to assume a consistent force when calculating work done on a charge over different paths or directions within the field.

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Most popular questions from this chapter

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?

An annulus with an inner radius of \(a\) and an outer radius of \(b\) has charge density \(\sigma\) and lies in the \(x y\) -plane with its center at the origin, as shown in Fig. \(\mathbf{P} 2 \mathbf{3 . 8 0}\). (a) Using the convention that the potential vanishes at infinity, determine the potential at all points on the \(z\) -axis. (b) Determine the electric field at all points on the \(z\) -axis by differentiating the potential. (c) Show that in the limit \(a \rightarrow 0, b \rightarrow \infty\) the electric field reproduces the result obtained in Example 22.7 for an infinite plane sheet of charge. (d) If \(a=5.00 \mathrm{~cm}, b=10.0 \mathrm{~cm}\) and the total charge on the annulus is \(1.00 \mu \mathrm{C}\), what is the potential at the origin? (e) If a particle with mass \(1.00 \mathrm{~g}\) (much less than the mass of the annulus) and charge \(1.00 \mu \mathrm{C}\) is placed at the origin and given the slightest nudge, it will be projected along the \(z\) -axis. In this case, what will be its ultimate speed?

A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) moves from the point \(x=0.150 \mathrm{~m}, y=0\) to the point \(x=0.250 \mathrm{~m}, y=0.250 \mathrm{~m} .\) How much work is done by the electric force on \(q_{2} ?\)

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

A proton and an alpha particle are released from rest when they are \(0.225 \mathrm{nm}\) apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

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