Download Problem Sheet 6

PHYS1022 Electricity and Magnetism
Problem Sheet 6 - workshop
1
A 100-pF capacitor and a 400-pF capacitor are charged to 1 kV and 2 kV
respectively. They are then disconnected from the voltage source and are connected
together, positive plate to positive plate, and negative plate to negative plate.
(a) Find the resulting potential difference across each capacitor.
(b) Find the energy lost when the connections are made.
2.
A uniformly charged thin ring has radius 15.0 cm and has total charge +24.0 nC. An
electron is placed on the ring’s axis a distance 30.0 cm from the centre of the ring and
is constrained to stay on the axis of the ring. The electron is then released from rest.
(a) Derive an expression for the potential along the axis due to the charged ring. Draw a
sketch of this function, and a sketch of the potential energy of the electron.
(b) Describe the subsequent motion of the electron by sketching its position along the
axis as a function of time.
(c) Find the speed of the electron when it reaches the centre of the ring.
3. Show that the electric field for an infinitely long, uniformly charged cylindrical shell of
radius R carrying a surface charge density σ is given by
Er = 0 for r < R
E 
r
R

for r > R

 r 2 r
0
0
where λ = 2πRσ is the charge per unit length on the shell.
1
Solutions
Q1 = C1 V1 = 10-10.103 = 10-7 C
Q2 = C2 V2 = 4x10-10. 2x103 = 8x10-7 C
After connection the system is a single capacitor with capacitance C1 + C2
And with Q = 9x10-7C
(a) V = Q/C = 9x10-7/5x10-10 = 1.8x103V
(b) Initial energy = ½ (C1 V12 + C2 V22) = 8.5x10-4 J
Final energy = ½ (C1 + C2) V2 = 8.1x10-4J
E = 4x10-5 J
1. Initially
2. All pieces of the ring are equidistant from the point and potential contribution from each
point round the ring just add…
(a)
so V = Q/4  0 r
Note U(x) = - e V(x)
(b) To begin all the energy is potential energy. The electron is attracted to the hoop
gaining kinetic energy. It will overshoot and by symmetry get to x=-30cm before
coming to rest… then oscillate
(c) Use conservation of energy: ½ m v2 = U = Q/4  0 ( 1/a – 1/(a+x)1/2)
v = 1.67 x 107 ms-1
3. Use a cylindrical Gaussian surface as shown.
The field must be radial and the same magnitude at any given r by symmetry. For r < R
the enclosed charge is zero
E =0
For r<R the enclosed charge is Q =  2  R L. Again using Gauss’ law we have
|E| 2  r L =  2  R L/0
|E| = R/ r 0
2