Download Homework Problem Set 3 Question 1 (1 point) 1. What is Gauss` Law

Homework Problem Set 3
Question 1 (1 point)
1. What is Gauss’ Law and in what cases is it useful for calculating electric fields?
Gauss’ Law:
Where Electric Field is constant over the whole surface that is CLOSED.
 E = electric field
 A = area
 Q = Charge
 0 = electric permittivity of space
*This is only useful for a closed, highly symmetrical system (such as planes of charge, lines
of charge, cylinders of charge, spheres of charge
Question 2 (3 points)
2. Two charged objects sitting on a horizontal surface. The first object is located at the origin with a
charge of +0.25 mC. The second object is located at (x,y) = (0,0.5m) with a charge of -0.25 mC.
What was the energy required to move both charges from an infinite distance away to the positions
they are in now? If the electric potential an infinite distance away is set to be 0V, what is the electric
potential at the point halfway between the two charges?
Work =
Work= Work = -
ds where s is some path opposite to r (which points outward):
dr
and since F = kq1q2/r2
kq1q2/r2dr in the direction of r
*Since the charges and coulombs constant do not change, they can be pulled
out of the integral
Work = -kq1q2 dr/r2
Work = kq1q2/r from a to 
Work = k q1q2/a
Work = (9 x 109Nm2/C2)(0.25mC)(-0.25mC) = -1125J (no work done-gives energy)
0.5m
Electric Potential :
kq1/r + kq2/r = 0
*Since this is halfway between 2 equal and opposite charges there is no potential
Question 3 (3 points)
3. An infinitely long rod sits horizontally along the x-axis with a continuous uniform charge
distribution such that the linear charge density λ is 0.015 C/m. At a point 0.35 m directly below the
rod what is the strength of the electric field? Use Gauss’ Law to solve.
 = 0.015 C/m
EdA
(for curved side only- since the angle between the field and normal force is 0
dot product becomes a scalar
the
EdA = EAcurved = E2rl = qin / o = l / o  E = /(2ro) = 2k/r
E= (2)(9x109 Nm2/C2)(0.015C/m)
0.35m
= 7.7 x 108 N/C or 7.7 x 108 V/m
Question 4 (3 points)
4. The electric potential around a single point charge can be expressed as V = kq/r
Derive Coulomb’s Law from this relation.

V=kq/r  Coulomb’s Law?
E = ∂V ∂V ∂V
∂x , ∂y , ∂z
E= ∂V 0 , 0
∂x,
*Where x is in the r direction, and y and z are
tangential to the circle of equipotential and
will both go to 0. So:
*Where x = r vector
∂V in the direction r = (d/dr)(kq/r) in the direction r = (kq)(d/dr)(1/r) in the direction r
∂r
E = -kq1 in the direction r and since F = q2E 
F = -kq1q2 in the direction r
r2
r2