Download Ph 213 – Challenging Problems (set3) Name: Due: July 16, 2013

Ph 213 – Challenging Problems (set3)
Due: July 16, 2013
Name:
Solution Key
1. Two parallel plates are 2.0 cm apart and the electric
field strength between them is 1.0 x 104N/C. An
electron is launched at 45o angle from the positive
plate. What is the maximum initial speed vo the
electron can have without hitting the negative plate?
Solution:
The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations
apply to the motion of the electron.
Visualize: The condition for the electron to not hit the negative plate is that its vertical velocity
should just become zero as the electron reaches the plate.
The force on the electron inside the capacitor is




 qE
F = ma = qE ⇒ a =
m
−(1.60 × 10−19 C)(1.0 × 104 N/C)
⇒ ay =
=
−1.756 × 1015 m/s 2
−31
9.11 × 10 kg
=
v0 y v0 sin 45°. Because the
=
v0 x v0 cos 45° and
The initial velocity v0 has two components:
electric field inside the capacitor is along the +y-direction, the electron has a negative acceleration
that reduces the vertical velocity. We require v1y = 0 m/s if it is not to hit the plate. Using
kinematics,
v12y = v02y + 2a y ( y1 − y0 ) ⇒ (0 m/s) 2 = v02y + 2a y ∆y
⇒ v0 y =
−2a y ∆y =
⇒ v0 =
−2(−1.756 × 1015 m/s 2 )(0.02 m) = 8.381 × 106 m/s
8.381 × 106 m/s
= 1.19 × 107 m/s
sin 45°
2. A sphere of radius R has a total charge Q throughout its volume. The volume charge density (C/m3)
within the sphere is ρ (r) = a/r2, where a is a constant to be determined. (see problem 27.57)
A) In terms of R and Q determine the value of constant a.
B) Use Gauss’s Law to find an expression for the electric field E inside the sphere for r << R.
C) Does your expression of the electric field found in part B, have the expected value at the surface
of the sphere (r = R)? Explain.
Solution:
A) Consider the spherical shell of radius r and thickness dr shown in the figure. The charge dq
within a small volume dV is
dq = ρdV =
a
(4πr 2 ) dr = 4π a dr
2
r
Integrating this expression to obtain the total charge in the sphere:
R
Q = ∫ dq = ∫ 4π a dr = 4π a R
0
⇒ a=
Q
4πR
B) Consider the spherical Gaussian surface at r < R shown in the figure. Gauss’s law applied to
this surface is
  Q
Q
Qenc
Φ E = ∫ E ⋅ dA = enc ⇒ E (4π r 2 ) = enc
⇒ E=
A
ε0
ε0
4πε 0 r 2
Using the results from part (A),
Qenc = ∫ dq = ∫ ρdV = ∫
r
0
E=
a
4π r 2 dr = 4π a r but
r2
Qenc
1
Qr
=
⋅
2
R 4πε 0 r 2
4πε 0 r

⇒ E=
a=
Q
4πR
⇒ Qenc =
Q
r
R
1
r
4πε 0 R r
Q
⋅
C) At r = R, the equation for the electric field obtained in part (B) simplifies to

Q

E=
⋅r
2
4πε 0 R
This is the same result as obtained in Example 27.3. The result was expected because a spherical
charge behaves, for r ≥ R, as if the entire charge were at the center.