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An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00 \(\mu\)s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

Short Answer

Expert verified
(a) The electric field is \(5.694 \times 10^{12}\, \text{N/C}\) downward. (b) Gravity can be ignored due to its negligible effect.

Step by step solution

01

Identify the Known Values

We have the distance \( d = 4.50 \text{ m} \) and the time \( t = 3.00 \times 10^{-6} \text{ s} \). The initial velocity \( v_0 = 0 \) because the electron starts from rest.
02

Calculate the Acceleration

Use the equation of motion for constant acceleration: \[ d = v_0 t + \frac{1}{2} a t^2 \]Substitute the known values:\[ 4.50 = 0 + \frac{1}{2} a (3.00 \times 10^{-6})^2 \]\[ 4.50 = \frac{1}{2} a (9 \times 10^{-12}) \]Solve for \( a \):\[ a = \frac{9.00}{9 \times 10^{-12}} = 1 \times 10^{12} \text{ m/s}^2 \]
03

Calculate the Electric Field Magnitude

Using the formula for force in an electric field \( F = ma \) and the charge of the electron \( e = 1.60 \times 10^{-19} \text{ C} \), we can equate it to \( F = eE \).\[ ma = eE \]\[ (9.11 \times 10^{-31} \text{ kg}) (1 \times 10^{12} \text{ m/s}^2) = (1.60 \times 10^{-19} \text{ C}) E \]Solve for \( E \):\[ E = \frac{9.11 \times 10^{-19}}{1.60 \times 10^{-19}} = 5.694 \times 10^{12} \text{ N/C} \]
04

Determine the Direction of the Electric Field

Since the electron is accelerated upward and the electric field direction is opposite to the acceleration for electrons (negative charge), the electric field must be directed downward.
05

Evaluate the Effect of Gravity

Calculate the gravitational force on the electron:\[ F_g = mg = (9.11 \times 10^{-31} \text{ kg})(9.81 \text{ m/s}^2) = 8.936 \times 10^{-30} \text{ N} \]Compare with the electric force:\[ F_E = eE = (1.60 \times 10^{-19} \text{ C})(5.694 \times 10^{12} \text{ N/C}) = 9.11 \times 10^{-7} \text{ N} \]Since \( F_E \gg F_g \), the effects of gravity are negligible in comparison.

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Key Concepts

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

Electron Motion
Electrons are tiny particles with a negative charge, and they obey the laws of physics just like larger objects do. When an electron is subjected to an electric field, it experiences a force that acts on it, pushing it towards a region of opposite charge. In this exercise, we see electron motion happening because of a uniform electric field. When the electron is released from rest, it begins to move because the electric field exerts a force on it. This force is not just static; it creates acceleration, meaning the electron speeds up or slows down depending on the field's direction. The equation of motion, which is a tool in physics to describe how objects move, helps us calculate the acceleration. Here, the electron travels 4.50 meters in 3 microseconds, which tells us it is accelerating rapidly.It's key to understand that because electrons have such a small mass, they accelerate very quickly under the influence of a strong force. The acceleration calculated from our example is a staggering \(1 \times 10^{12} \text{ m/s}^2\). This shows just how sensitive electrons are to forces like those from electric fields.
Uniform Electric Field
A uniform electric field is one where the strength and direction of the field are consistent throughout. In simple terms, it means that any charged particle, like an electron, will feel the exact same force no matter where it is in the field. This property simplifies many physics calculations because you don't have to worry about changes in the electrical force over space.In our example, this consistency allows us to determine the exact acceleration of the electron because we know the electric field's magnitude and direction don't change. The formula that connects these concepts is \(F = eE\), where \(e\) is the electron's charge, and \(E\) is the electric field strength. For an electron, the field direction is typically perpendicular to its path and for our case, it's downward as calculated.Understanding the electric field's direction helps comprehend electron behavior. Since electrons are negatively charged, they're accelerated in the opposite direction of the field in contrast to positive charges, which move in the field's direction.
Gravitational Force Negligibility
In an earth-bound environment, both gravitational and electric forces can act on charged particles like electrons. However, due to the electron's small mass, the gravitational force it experiences is minimal. Comparatively, the electric force is significantly larger for an electron in a strong electric field. When we compute these forces, the gravitational force \(F_g\) turns out to be \(8.936 \times 10^{-30} \text{ N}\), which is tiny when put next to the electric force \(F_E = 9.11 \times 10^{-7} \text{ N}\).This enormous difference—where \(F_E\) is millions of times stronger than \(F_g\)—justifies our choice to ignore gravity when accounting for electron motion in a uniform electric field. The impact of gravitational force becomes negligible and doesn't significantly affect the electron's path or acceleration compared to the dominant electric force.

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Most popular questions from this chapter

In a rectangular coordinate system a positive point charge \(q = 6.00 \times \space 10^{-9} C\) is placed at the point \(x = +0.150 m, y = 0,\) and an identical point charge is placed at \(x = -0.150 m, y = 0.\) Find the \(x\)- and \(y\)-components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x =\) 0.300 m, \(y =\) 0; (c) \(x =\) 0.150 m, \(y = -\)0.400 m; (d) \(x = 0, \)y =$ 0.200 m.

If a proton and an electron are released when they are 2.0 x 10\(^{-10}\) m apart (a typical atomic distance), find the initial acceleration of each particle.

A ring-shaped conductor with radius \(a =\) 2.50 cm has a total positive charge \(Q = +\)0.125 nC uniformly distributed around it (see Fig. 21.23). The center of the ring is at the origin of coordinates \(O\). (a) What is the electric field (magnitude and direction) at point \(P\), which is on the \(x\)-axis at \(x =\) 40.0 cm? (b) A point charge \(q = -2.50 \space \mu\)C is placed at \(P\). What are the magnitude and direction of the force exerted by the charge \(q\) \(on\) the ring?

Two thin rods of length \(L\) lie along the \(x\)-axis, one between \(x = \frac{1}{2} a\) and \(x = \frac{1}{2} a + L\) and the other between \(x = -\frac{1}{2} a\) and \(x = -\frac{1}{2} a - L\). Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive x-axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$F = {Q^2 \over 4\pi\epsilon_0 L^2} ln [ {(a + L)^2 \over a(a + 2L)} ]$$ (c) Show that if \(a\) \(\gg\) \(L\), the magnitude of this force reduces to \(F = Q^2/4\pi\epsilon_0 a^2\). (\(Hint\): Use the expansion ln \((1 + z) = z - \frac{1}{2} z^2 + \frac{1}{3} z^3 - \cdot\cdot\cdot\), valid for \(\mid z \mid\ll1\). Carry \(all\) expansions to at least order \(L^2/a^2.\)) Interpret this result.

A small sphere with mass \(m\) carries a positive charge \(q\) and is attached to one end of a silk fiber of length \(L\). The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density \(\sigma\). Show that when the sphere is in equilibrium, the fiber makes an angle equal to arctan (\(q\sigma/2mg\epsilon_0\)) with the vertical sheet.

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