Chapter 24, Problem 038 The figure shows a thin plastic rod of length L 11.1 cm

Question

Chapter 24, Problem 038 The figure shows a thin plastic rod of length L 11.1 cm and uniform charge 16.6 fC. (a) n terms of distance d, charge density and Eo fin an expression for the electric potential at point P d (b) Next, substitute variable x for d and find an expression for the magnitude of the component Ex of the electric field at P n terms of d and other variables). (c) What is the direction of E relative to the positive direction of the x axis? (d) What is the value of Ex at P1 for x d 3.48 cm? (e) From the symmetry in figure determine Ey at P1. k-d ck here to enter or edit your answer (a)V

Explanation / Answer

a.

consider a small length of dx at x.

charge on this small length=dq=lambda*dx

where lambda=linear charge density

this is at a distance of d+x from P1.

so electric potential at P1 due to dq=dV=k*dq/(d+x)=k*lambda*dx/(d+x)

where k=1/(4*pi*epsilon)

integrating dV over x=0 to x=L

we get total potential =k*lambda*ln(d+x)

putting the limits, potential at P1=lambda*(k*ln(d+L)-k*ln(d))

=k*lambda*ln((d+L)/d)

using k=1/(4*pi*epsilon)

total potential at P1=lambda*(1/(4*pi*epsilon))*ln((d+L)/d)

part b:

electric field at P due to dq=k*dq/(d+x)^2=k*lambda*dx/(d+x)^2

integrating from x=0 to x=L,

total electric field at P=Ex=-k*lambda/(d+x)

using the limits,

Ex=lambda*((-k/(d+L))-(-k/(d)))

=k*lambda*((1/d)-(1/(d+L))

=k*lambda*L/(d*(d+L))

using k=1/(4*pi*epsilon)

Ex=lambda*(1/4*pi*epsilon)*L/(d*(d+L))

part c:

as charge is positive, field is directed away from it.

so Ex is along -ve x axis.

part d:

using d=3.48 cm=0.0348 m

L=11.1 cm=0.111 m

lambda=16.6 fC/L=16.6*10^(-15)/0.111=1.4955*10^(-13) C/m
Ex=0.02944 N/C

part e:

Ey=0

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.