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Chapter 21: Problem 78

A small object with mass \(m,\) charge \(q,\) and initial speed \(v_{0}=5.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) is projected into a uniform electric field between two parallel metal plates of length \(26.0 \mathrm{~cm}\) (Fig. \(\mathrm{P} 21.78\) ). The electric field between the plates is directed downward and has magnitude \(E=800 \mathrm{~N} / \mathrm{C}\). Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance \(d=1.25 \mathrm{~cm}\) from its original direction of motion and reaches a collecting plate that is \(56.0 \mathrm{~cm}\) from the edge of the parallel plates. Ignore gravity and air resistance. Calculate the object's charge-to-mass ratio, \(q / m\).

Short Answer

Expert verified
Plugging in the values: \(q/m = 0.0125m / (0.5 \cdot 800N/C \cdot ((0.26m / 5 \cdot 10^{3}m/s) + (0.3m / 5 \cdot 10^{3}m/s))^2) = 1.58 \times 10^{-7} C/kg\)

Step by step solution

01

Find Acceleration

The object experiences a force due to the electric field. This force is given by \(F = qE\) where \(E\) is the electric field. The acceleration \(a\) of the object in the field can be found using Newton's second law, \(F = ma\), which yields \(a = qE/m\). Given that the magnitude of the electric field \(E = 800 \, N/C\), the acceleration \(a\) is expressed as \(a = qE/m = 800q/m \, m/s^2\).
02

Determine Vertical Distance in Field

As the object travels through the field, it is deflected downwards. The time it takes for the object to travel the length of the plates is \(t = L/v_0\), where \(L = 26 \, cm = 0.26 \, m\) is the length of the plates and \(v_0 = 5.00 \times 10^{3} \, m/s\) is the initial speed of the object. The distance travelled by the object in this time due to the downward acceleration is given by the equation of motion \(d1 = 0.5 \cdot a \cdot t^2 = 0.5 \cdot 800q/m \cdot (0.26/5.00 \times 10^{3})^2\).
03

Determine Total Vertical Distance

After passing through the field, the object continues to move downward with the same vertical velocity it attained in the field. The time it takes to travel the remaining horizontal distance to the collecting plate is \(t' = D/v_0\), where \(D = 56 \, cm - 26 \, cm = 30 \, cm = 0.3 \, m\) is the horizontal distance of the collecting plate from the edge of the plates. The vertical distance covered in this time is \(d2 = v_{fy} \cdot t' = a \cdot t \cdot t'\). Therefore, the total vertical distance \(d\), being the sum of \(d1\) and \(d2\), that the object is deflected is given by \(d = d1 + d2 = 1.25 \, cm = 0.0125 \, m\).
04

Solve for Charge-To-Mass Ratio

Solving the equation obtained in Step 3 for \(q/m\) yields \(q/m = d/(0.5 \cdot E \cdot (t + t')^2)\)

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Key Concepts

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

Uniform Electric Field
A uniform electric field is one where the electric force on a charge is constant in both magnitude and direction throughout the entire region. Imagine being in a room where the temperature is the same no matter where you stand鈥攖hat's a bit like a uniform electric field for charged particles.

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Most popular questions from this chapter

An American penny is \(97.5 \%\) zinc and \(2.5 \%\) copper and has a mass of \(2.5 \mathrm{~g}\). (a) Use the approximation that a penny is pure zinc, which has an atomic mass of \(65.38 \mathrm{~g} / \mathrm{mol},\) to estimate the number of electrons in a penny. (Each zinc atom has 30 electrons.) (b) Estimate the net charge on all of the electrons in one penny. (c) The net positive charge on all of the protons in a penny has the same magnitude as the charge on the electrons. Estimate the force on either of two objects with this net magnitude of charge if the objects are separated by \(2 \mathrm{~cm}\). (d) Estimate the number of leaves on an oak tree that is 60 feet tall. (e) Imagine a forest filled with such trees, arranged in a square lattice, each \(10 \mathrm{~m}\) distant from its neighbors. Estimate how large such a forest would need to be to include as many leaves as there are electrons in one penny. (f) How does that area compare to the surface area of the earth?

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

A semicircle of radius \(a\) is in the first and second quadrants, with the center of curvature at the origin. Positive charge \(+Q\) is distributed uniformly around the left half of the semicircle, and negative charge \(-Q\) is distributed uniformly around the right half of the semicircle (Fig. P21.84). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?

A straight, nonconducting plastic wire \(8.50 \mathrm{~cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point \(6.00 \mathrm{~cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point \(6.00 \mathrm{~cm}\) directly above its center.

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?
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