124. The charge per unit length on the thin rod shown below is A. What is the electric force on the point charge q? Solve this problem by first considering the electric force dF on q due to a small segment dæ of the rod, which contains charge Adz. Then, find the net force by integrating dF over the length of the rod.

esc The nucleus of a 125 Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q = +54e. R 2 R F2 W # 3 80 E $ 4 R Part A What is the electric force on a proton 2.6 fm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point charge. Express your answer with the appropriate units. F= 8.64 Part B μà 1 Submit Previous Answers Request Answer % 5 X Incorrect; Try Again; 4 attempts remaining F5 What is the proton's acceleration? Express your answer with the appropriate units. Value T → HA N 6 C FIC ? Units MacBook Air Y & 7 U * 8 FB ( 9 F9 0 0 P + Review

Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). a. Find the electric charge due to the protons in the dust. Calculate the charge due to only the protons.

A thin, insulating rod of length Icarries a linear charge density^(x) that varies with distance according to (x)=Ax (in SI units). The location of the origin is shown in the figure. A point charge q is located a distance I from the end of the rod as shown. a. What are the SI units of the constant A? b. Find the force that the rod exerts on q.

A single isolated, large conducting plate has acharge per unit area σ on its surface. Because the plate is a conductor, the electric field at its surface is perpendicular to the surface and has magnitude E = σ/εo a.The field from a large, uniformly charged sheet with charge per unit area σ has magnitude E = σ/2εo. Why is there a difference? b.Regard the charge distribution on the conducting plate as two sheets of charge (one on each surface), each with charge per unit area σ. Find the electric field inside and outside the plate.

A pair of parallel conducting plates are given charges of equal magnitude but opposite sign to create a uniform electric field with magnitude 38 N/C. A rectangular surface with dimensions 4.2cm×1.2cm is located in the gap between the parallel plates. a. What is the magnitude of the electric flux, in newton squared meters per coulomb, through the rectangular surface if it is parallel to the charged plates?  b. What is the magnitude of the electric flux, in newton squared meters per coulomb, through the rectangular surface if it is perpendicular to the charged plates?  c. What is the magnitude of the electric flux, in newton squared meters per coulomb, through the rectangular surface if the angle between its normal and the electric field is 25∘?

A negative line of charge is on the negative y axis with its top end at the origin as shown in the figure below. The charge is uniformly distributed along the line. Point P is on the positive y axiS as shown in the figure. y The the length of the line of charge is 3.17 m, the total charge on the line is -787.9 nC and point P is 3.15 m from the origin. Find the magnitude of the electric field at point P due to the line of charge. (in N/C) OA: 3.15×10²| OB: 3.56×102| 0C: 4.02×102| OD: 4.54×102| OE: 5.13×102|| OF: 5.80×102|| OG: 6.56×102|| OH: 7.41×102 Submit Answer Tries 0/99

Consider a charge configuration comprising two point sources, one carrying a positive charge and the other a negative charge. The magnitude of the negative charge is three times as large as the positive charge. Construct a careful sketch of the electric field around the configuration.