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Unformatted text preview: 25.3.1 Potential and Work Done Three point charges are placed at equal distance from 0.
Find the potential at 0. Choose one: V at O —kq 37a} kq a a a Hints: V=ZiV}, with Vi=in/ri Extra: Show that the work done in bringing another charge
q1 from inﬁnity to O is qulla. k
_q.
a Explanation: At 0, VEV0= Extra: Work from inﬁnity to the origin, qu1
W=PE0—PEOO=q1V0=T_ 25.3.3 Three charges at comers an equilateral triangle
9J3, q Three charges are located at the vertices of an equilateral
triangle, see sketch. Excluding the charge at A, determine the
direction of electric ﬁeld vector and the potential at A. Choose one:
1 2 3 4 
dir. of EA \ \ V\ V\ ‘
VA 0 —2kq/a 0 +2kq/a
Hint: To determine the direction of E, you should ﬁrst sketch a
vector diagram. Extra: Show the work required to set up the conﬁguration
including the charge at A is —kq2/a.
Explanation: At A, the vector diagram of E AB+E AC is given VA=VAB+VAC= kq/a +kq/a=0. Ans=3.
Extra: Total energy is given by:
U AB+UBC+UC A = k(q)2/a  qu/a  qu/a =  qula. 25.3.2 Potential energy, work done, V and E O A B
 Hq Q A positive point charge Q is located at 0. Given the work
in bringing another positive charge q from inﬁnity to the point A is Woo_)A=lJ. Find the work required in bringing the same q from
inﬁnity to the point B, where 0B=2a, with a=OA. Choose one: 1 2
0.25 0.5 Woo—>B(in 1) Hint: W00_,A = PEA=qu/a.
Extra: Ifa= 2m, q=lC show that VA=1V and EA=0.5V/m. Explanation: Wm_,3= PEB=qu/(2a)=(qu/a)/2. Here qu/a=l J, since it is the potential energy at A. So
Woo_,B= 1/2=0.5J. Extra: VA=PEA/q=1/1=1V. E A=qu/a2=(qu/a)la=1/2=0.5V/m.
A comment on units: V/m=j/(mC)=Nm/(mC)=N/C. 25.4.1 Potential V and components of E Given: Q1=Q>0, Q2: —3Q. AP=BP=r.
Find: The potential V, at the point P. 1 2 3 4
VatP Q _E _ﬂ &
r r I‘ l‘ Him; V= EllVi, with =kqJ‘ri
Extra: Find the direction of Ex at P by inspection. Explanation:
V kQ1+kQ2 kQ_3kQ _2kQ
— r r — r r — 1‘ Extra: By inspection, at the point P the projections of
both the electric field vectors El and E2 are along the
positive x—axis. So the vector sum E=El+E2 has a
projection along the positive x—axis. A challenge: Please verify, EX ax 25.4.2 Potentials at different points near a dipole The field pattern and a pair of equipotential curves of a
dipole are shown in the sketch. Compare potentials at A, B and C. Choose one: Hint: Natural tendency for a positive charge is to go from
a high potential point to a low potential point. Explanation: By inspection on the sketch below,
VA>V~ , VA, = VB :0 , and VB>VC. Ans = 1. 25.5.2 Uniformly charged sphere Given a uniformly charged sphere with a total charge Q
and a radius R. It can be shown that the electric ﬁeld E=pr/(380). The potential difference between A, where
OAq<R and B, the point along the same radial line on the surface of the sphere, is given by:
1 2 4.2% tam 3
E(Rr) V(r)V(R) Extra: Show the potential difference
V(r)V(R)=[kQ/(2R)][1(f/R)2] Explanation: AV: —EAs. For the present case E depends
on r. So the potential difference must be evaluated through
an integral. By inspection, Ans=1 gives the desired
positive potential difference. Extra: Using E=pr/(3so), the integral gives
v(r)V(R)=w<6e,)1[R242] (1) [p/(680)]= [Q/(4nR3/3)]/(6so)= Q/(81IR3 so) = kQ/2R3. (2)
From (2) & (1), V(r)V(R)=[kQ/(2R)][1(r/R)2]. 25.5.1 Two conducting spheres which are far apart Two Conducting spheres which are far apart and are kQ kQ
connected by awire. Assume V ” —1,V ” —2.
1 r1 2 r2 ICompare the charges on the two spheres, i.e. Q1 vs Q2 Choose one:
Q1>Q2 Q1=Q2 Q1<Q2 Hint: A conductor is an equipotential body, i.e. V1: V2. Extra: Being far apart, the ﬁelds are expected to be
. le sz
given by: E1~—2, EZz—Z. Show that E1>E2 .
r1 r2
Explanation:
le sz Q1 r1
V1=—=V2=—. So —=— <1, or Q1<Q2. r1 r Q2 r2 Q Q1: Q2=Q Given two conducting spheres which are separated by a large distance. One is smaller than the other, i.e. rl<r2. Each has a positive charge Q on it. Being far apart, the
potentials at #1 and at #2 are assumed to be approximately given by:
Vlzg, and V2 sz . IE1 1r2
We connect them by a conducting wire. Describe the
direction of the “apparent ﬂow” of positive charges. (Technically speaking, electrons are the ones which are
ﬂowing in the opposite direction.) Positive charges Positive charges No flow
ﬂow from flow from due to same charges
#1 to #2 #2 to #1 on both bodies Hint: Natural tendency for positive charges is to ﬂow
from a high potential to a low potential. Explanation: Since r1< r2, from the equation above V1>
V2. The flow of positive charges is from #1 to #2. We mention once again, what really happens is that the
negative charges ﬂow from #2 to #1. 25.6.2 Potential due to two conducting spherical shells Given two thin concentric conducting spherical shells
with charges Q and —Q on the inner and the outer shells
respectively. Find: V at P in region I, which is a distance r from 0.
Choose one: Hint: Us the superposition principle. Extra: Show the potential in region 11 at a distance r
from O is given by V=ler — kab. What is the potential
at O? Explanation: In region I, using the superposition
principle, V: Vim, + Vomer : (kQ/r) — (kar) = 0.
Extra: At r=a, Va=(kQ/a) —(kQ/b). But within the
conducting shell, i.e. in region [1], E20. In turn, V
should be constant throughout, which includes the point 0. In other words, at O: V0=Va=(kQ:’a) (kQ:’b). 25.6.3 Potential due to two conducting spherical shells Given two thin concentric conducting spherical shells with
charge Q on the inner shell (with radius a) and charge —Q
on the outer shell (with radius b ). Find: V — Vb.
a kQ/b kQ/a + kQ/b kQ/a — kQ/b Hint: Use the superposition principle. First show at P, the
potentials is 0, or Vb = 0. Extra: What is the potential at O ? Explanation: In region ll , V: (kQ/r) — (kQ/b). So at r=a,
V=(kQ/a)  (kQ/b) VaVb = (kQ/a)  (kQ/b) Extra: 0 is in region [11. A conductor is an equipotential
body, so Vo =Va = (kQ/a) — (kQ/b). ...
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This note was uploaded on 02/04/2012 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics

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