Physics Challenge for
Teachers and Students
Boris Korsunsky, Column Editor
Weston High School, Weston, MA 02493
korsunbo@post.harvard.edu
Solution to May 2011 Challenge
Therefore, the resultant charges are: w A Late Charge
Two capacitors are connected by the ideal wires forming a square
with side d. The central wire connects the midpoints of the opposite sides of the “square” as shown. The circuit is placed in the
magnetic field, directed into the page, that changes with time as
B(t) = kt, where k is a constant coefficient. After some time, the
central wire is cut and the magnetic field is then kept constant.
Find the charges on the capacitors after the equilibrium has been
reached. The capacitances C1 and C2 are given.
d
C1
Q1 = C1V =
(C2 − C1 ) C1kd 2
(C1 + C2 ) 2
and Q2 = C2V =
(C2 − C1 ) C2 kd 2
.
(C1 + C2 ) 2
A comment on signs: If C2 > C1 then, as shown in the
figure below, both upper capacitor plates will be positively
charged; otherwise, for C2 < C1 the upper plates will be
negatively charged.
C2
We received quite a few solutions from all over the world
to our May Challenge, A Late Charge. Here is one of the
many correct solutions:
(Submitted by Bartley L. Cardon, South Portland, ME)
Solution: By Faraday’s law, an emf, E=−
2
dΦ
dB
d (kt )
kd
,
= −A
= −(d ⋅ d )
=−
2
2
dt
dt
dt
is induced in each half-loop. For the duration that the
B-field acts, E is the potential across each capacitor. Hence,
each capacitor C1 and C2 acquires a charge given by the
general formula Q = CE , or Q1 = –C1E and Q2 = C2E. The
different signs on the Q’s are the result of the current in the
opposite directions to ensure that Lenz’s law is obeyed.
The central wire is now cut and the B-field held constant. With the B-field constant, there is no changing flux
through the wire loop connecting the two capacitors, and
E disappears. However, charges Q1 and Q2 must redistribute to bring both capacitors, now in parallel, to the same
potential V. Treating the parallel capacitors as one, with
capacitance C = C1 + C2 and a total charge
Q = Q1 + Q2 = E (C2 − C1 ) =
we find
Q (C − C1 ) kd 2
V= = 2
.
C (C1 + C2 ) 2
kd 2
(C2 − C1 ),
2
We are pleased to recognize the following contributors:
Mehmet Nuri Akinsi (İstanbul Technical University, Istanbul,
Turkey)
Sharmila Balamurugan, student (Women’s Christian College,
Chennai, India)
Hratch Barsoumian (Haigazian University, Beirut, Lebanon)
Phil Cahill (Lockheed Martin Corp., North Yorkshire, UK)
Juan Carlos Criado, student (Universidad de Granada, Spain)
Don Easton (Lacombe, Alberta, Canada)
Fernando Ferreira (Univ. da Beira Interior, Covilhã, Portugal)
Fredrick P. Gram (Cuyahoga Comm. College, Cleveland, OH)
Norge Cruz Hernández (University of Seville, Seville, Spain)
Gerald E. Hite (TAMUG, Galveston, TX)
Art Hovey (Milford, CT)
José Ignacio Íñiguez de la Torre (Universidad de Salamanca, Salamanca, Spain)
Joep Jongen (Hogeschool, Utrecht, Netherlands)
Carl E. Mungan (U. S. Naval Academy, Annapolis, MD)
Michael Rapport (Anne Arundel Comm. College, Arnold, MD)
Clint Sprott (University of Wisconsin – Madison, WI)
Many thanks to all contributors and we look forward to
hearing from you!