Concept explainers
Often we have distributions of charge for which integrating to find the electric field may not be possible in practice. In such cases, we may be able to get a good approximate solution by dividing the distribution into small but finite particles and taking the vector sum of the contributions of each. To see how this might work, consider a very thin rod of length L = 16 cm with uniform linear charge density λ = 50.0 nC/m. Estimate the magnitude of the electric field at a point P a distance d = 8.0 cm from the end of the rod by dividing it into n segments of equal length as illustrated in Figure P24.21 for n = 4. Treat each segment as a particle whose distance from point P is measured from its center. Find estimates of EP for n = 1, 2, 4, and 8 segments.
FIGURE P24.21
The magnitudes of electric fields at P for the segments
Answer to Problem 21PQ
The magnitude of electric fields
Explanation of Solution
Write the expression to calculate the electric field.
Here,
Write the expression to calculate the charge in each segment.
Here,
Substitute the above equation in the expression for
Write the expression to calculate
Here, d is the distance of the point P from the end of the rod.
Write the expression to calculate
Write the expression to calculate
Write the expression to calculate
Substitute the equations (II), (III), (IV) and (V) in (I) to rewrite.
Conclusion:
Substitute
Similarly, by following the same concepts the electric field for
Therefore, the magnitude of electric fields
Want to see more full solutions like this?
Chapter 24 Solutions
Physics for Scientists and Engineers: Foundations and Connections
- Consider the electric field of a point charge.At point A the electric field strength is 9 V/m, at point B the electric field strength is 36 V/m. We measure the strength of the electric eld at the midpoint of the interval AB. What can be the measured value?arrow_forwardConsider two thin disks, of negligible thickness, of radius R oriented perpendicular to the x axis such that the x axis runs through the center of each disk. The disk centered at x=0 has positive charge density η, and the disk centered at x=a has negative charge density −η, where the charge density is charge per unit area. What is the magnitude E of the electric field at the point on the x axis with x coordinate a/2? Express your answer in terms of η, R, a, and the permittivity of free space ϵ0.arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.arrow_forward
- Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Calculating the antiderivative or indefinite integral , Vab = (-αa0e-r/a0 + β + b0 ) By definition, the capacitance C is related to the charge and potential difference by: C = / Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q / ( (e-rb/a0 - e-ra/a0) + β ln() + b0 () )arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Calculating the antiderivative or indefinite integral , Vab = (-αa0e-r/a0 + β + b0 ) By definition, the capacitance C is related to the charge and potential difference by: C = / Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q / ( (e-rb/a0 - e-ra/a0) + β ln() + b0 () )arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by:arrow_forward
- Find the electric field at the origin of the x,y-plane for charge distributions (a) and (b), see the figures shown below. The field is produced (a) by a thin half-circle with a radius of 15 cm and the linear charge density K-10 pc/cm and (b) by a thin quarter-circle with the same radius and the linear charge density K = -10 pc/cm. K>0 (a) For the charge distribution (a): The x-component of Ea. Ea,x= The y-component of Ea, Ea,y= For the charge distribution (b): The x-component of Eb, Eb,x- The y-component of Eb, Eb,y = Units N/C Units N/C Units N/C Units N/C K<0 (b)arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = aer/ao + B/r + bo %3D where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Vab = | S"Edr= - [ *Edr Calculating the antiderivative or indefinite integral, Vab = (-aage-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e-rb/ao - eralao) + B In( ) + bo ( ))arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = aer/ao + B/r + bo %| where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Va Edr= Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e-"b/ao - era/ao) + B In( ) + bo ( ))arrow_forward
- QUESTION 1 Problem: An infinitely long cylindrical conductor has radius R and uniform surface charge density o. In terms of R and o, what is the charge per unit length A for the cylinder? Answer: A = 2arrow_forwardA cylinder of length L=5m has a radius R=2 cm and linear charge density 2=300 µC/m. Although the linear charge density is a constant through the cylinder, the charge density within the cylinder changes with r. Within the cylinder, the charge density of the cylinder varies with radius as a function p( r) =p.r/R. Here R is the radius of the cylinder and R=2 cm and p, is just a constant that you need to determine. b. Find the constant po in terms of R and 2. Then plug in values of R and 1. to find the value for the constant p. c. Assuming that L>>R, use Gauss's law to find out the electric field E inside the cylinder (rR) in terms of 1. and R. d. Based on your result from problem c, find the electric field E at r=1cm and r=4cm.arrow_forwardA flat wire of infinite length has width W and charge density sigma. If we align the wire so that it is everywhere at y=0 and extends in x from x=d to x=d+L, then calculate the electric field at x=0. Hint: Treat the wire as an the sum of the contributions of many "long wires" where the linear charge density satisfies: lambda=sigma*L.arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning