A thin rod of length L = 0.5 m has a non-uniform linear charge density ฮป(x) = ฮปโx, where ฮปโ = 4.0 ร 10โปโถ C/mยฒ and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.
๐ก Show Solution
Given:
L = 0.5 m
ฮป(x) = ฮปโx where ฮปโ = 4.0 ร 10โปโถ C/mยฒ
d = 0.2 m
k = 8.99 ร 10โน Nยทmยฒ/Cยฒ
Setup:
For a small element dx at position x from the origin, the charge is:
Coulomb's law, electric field from continuous charge distributions
How can I study Electric Field and Coulomb's Law effectively?โพ
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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=
kr2q1โq2โโr^=
4ฯฯต0โ1โr2q1โq2โโr^
where:
k=8.99ร109 Nยทmยฒ/Cยฒ
ฯต0โ=8.85ร10โ12 Cยฒ/(Nยทmยฒ) (permittivity of free space)
Vector form:F12โ=kr122โq1โq2โโr^12โ
Electric Field
Electric field due to point charge:
E=kr2qโr^
Definition:E=q0โFโ
where q0โ is test charge.
Superposition principle:Etotalโ=โiโEiโ
Continuous Charge Distributions
For continuous distribution with charge density ฯ:
E=kโซr2dqโr^
Linear Charge Density
Charge per unit length: ฮป=dq/dl
dE=kr2ฮปdlโr^
Surface Charge Density
Charge per unit area: ฯ=dq/dA
dE=kr2ฯdAโr^
Volume Charge Density
Charge per unit volume: ฯ=dq/dV
dE=kr2ฯdVโr^
Example: Infinite Line of Charge
Uniform line charge density ฮป, find field at distance r:
By symmetry, field is radial. Consider element at distance z:
dExโ=(r2+z2)kฮปdzโr2+z2โrโ
E=โซโโโโ(r2+z2)3/2kฮปrdzโ
Using z=rtanฮธ:
E=r2kฮปโ=2ฯฯต0โrฮปโ
Example: Ring of Charge
Ring of radius R, total charge Q, find field on axis at distance x:
Exโ=(x2+R2)3/2kQxโ
At center (x=0): E=0 (by symmetry)
Far from ring (xโซR): EโkQ/x2 (point charge)
Example: Disk of Charge
Uniform surface charge density ฯ, radius R, field on axis:
Answer: The electric field at point P is 1.94 ร 10โด N/C pointing away from the rod.
2Problem 2medium
โ Question:
A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 ร 10โปโธ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?
๐ก Show Solution
Given:
R = 0.1 m
Q = 2.0 ร 10โปโธ C = 20 nC
k = 8.99 ร 10โน Nยทmยฒ/Cยฒ
Formula for field on axis:Exโ=(x2+R2)3/2kQxโ
(a) Field at x = 0.05 m:
E=[(0.05)2+(0.1)
E=[(0.0025+0.01)]3/2
E=1.40ร10โ30.899โ=
(b) Maximum field position:
To find maximum, take derivative and set equal to zero:
dxdExโโ
dxdE
=kQ(x2+R2
R2โ2x2=0
x=2โ
(c) Maximum field strength:
Emaxโ=[(R/
=[R2/2+R
=R22
=33
=0.052359.6โ=6.91ร102ย N/C
Answers:
(a) E = 642 N/C
(b) x = 7.07 cm = R/โ2
(c) E_max = 691 N/C
3Problem 3medium
โ Question:
An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 ร 10โปยนยฒ Cยทm. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.
๐ก Show Solution
Given:
d = 2.0 ร 10โปยณ m = 2.0 mm
p = 5.0 ร 10โปยนยฒ Cยทm
r = 0.1 m = 10 cm
k = 8.99 ร 10โน Nยทmยฒ/Cยฒ
(a) Magnitude of each charge:
Dipole moment: p = qd
q=dpโ=2.0ร10โ3
(b) Field on perpendicular bisector (far field, r >> d):
Eperpโ=r3
E=(0.1)3(8.99ร10
E=10โ34.495ร10
Direction: From +q toward -q (perpendicular to dipole axis)
(c) Field on axis (far field, r >> d):
Eaxisโ=r3
E=(0.1)32(8.99ร10
E=10โ38.99ร10
Direction: Away from dipole (from -q toward +q) for points on the side of +q
Note: The axial field is exactly twice the perpendicular field at the same distance:
Eaxisโ=2Eperpโ
Answers:
(a) q = 2.5 nC
(b) E = 45.0 N/C (perpendicular to axis)
(c) E = 89.9 N/C (along axis)**
โพ
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.