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Chapter 16: Problem 30

An electron traveling horizontally from west to east enters a region where a uniform electric field is directed upward. What is the direction of the electric force exerted on the electron once it has entered the field?

Short Answer

Expert verified
Answer: The electric force exerted on the electron is in the southeast direction (downward and eastward).

Step by step solution

01

Analyze the electric field and electron charge

The electric field is directed upward. Since the electron has a negative charge, the electric force acting on it will be in the opposite direction of the electric field. Therefore, the electric force on the electron will be directed downward.
02

Combine directions of electron movement and electric force

The electron is moving horizontally from west to east, and the direction of the electric force is downward. Combining these two directions, the overall direction of the electric force exerted on the electron will be southeast (downward and eastward).
03

Conclusion

The electric force exerted on the electron, once it has entered the field, is in the southeast direction (downward and eastward). This is due to the combination of the electron's horizontal movement from west to east and the downward direction of the electric force, which is opposite to the upward-directed electric field.

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Most popular questions from this chapter

Three equal charges are placed on three corners of a square. If the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{b}}\) has magnitude \(F_{\mathrm{ba}}\) and the force that \(Q_{\mathrm{a}}\) exerts on \(Q_{\mathrm{c}}\) has magnitude \(F_{\mathrm{ca}},\) what is the ratio of \(F_{\mathrm{ca}}\) to $F_{\mathrm{ba}} ?$
A positively charged rod is brought near two uncharged conducting spheres of the same size that are initially touching each other (diagram a). The spheres are moved apart and then the charged rod is removed (diagram b). (a) What is the sign of the net charge on sphere 1 in diagram b? (b) In comparison with the charge on sphere \(1,\) how much and what sign of charge is on sphere \(2 ?\)
Use Gauss's law to derive an expression for the electric field outside the thin spherical shell of Conceptual Example 16.8
In this problem, you can show from Coulomb's law that the constant of proportionality in Gauss's law must be \(1 / \epsilon_{0} .\) Imagine a sphere with its center at a point charge q. (a) Write an expression for the electric flux in terms of the field strength \(E\) and the radius \(r\) of the sphere. [Hint: The field strength \(E\) is the same everywhere on the sphere and the field lines cross the sphere perpendicular to its surface.] (b) Use Gauss's law in the form \(\Phi_{\mathrm{E}}=c q(\text { where } c\) is the constant of proportionality) and the electric field strength given by Coulomb's law to show that $c=1 / \epsilon_{0}$
Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=2 d\) (point \(S\) )?
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