C H A PT E R
Cu RRENT ELECTRICITY
R
SI
3.1 CURRENT ELECTRICITY If the current is steady i.e., the rate of flow of charge
1. What is current electricity ? does not change with time, then
Current electricity. In chapters 1 and 2, we studied 1= Q
IT
the phenomena associated with the electric charges at t
rest. The physics of charges at rest is called . Electric charge
electrostatics or static electricity. We shall now study or EIectnc current = ------"'-
Time
the motion or dynamics of charges. As the term current
implies some sort of motion, so the motion of electric where Q is the charge that flows across the given area
charges constitutes an electric current. in time t.
H
The study of electric charges in motion is called current Lightning, which is the flow of electric charge
electrici ty. between two clouds or from a cloud to the earth, is an
example of a transient current (a current of short
3.2 ELECTRIC CURRENT duration). But the charges flow in a steady manner in
devices like a torch, cell-driven clock, transistor radios,
O
2. Define electric current. hearing aids, etc.
El~tric current. If two bodies charged to different 3. Give the 51 unit of current.
poten .als are connected together by means of a
SI unit of current is ampere. If one coulomb of charge
conduc ing wire, charges begin to flow from one body
crosses an area in one second, then the current through that
to another. The charges continue to flow till the
area is one ampere (A).
M
potentials of the two bodies become equaL
1coulomb
Theflow of electric charges through a conductor constitutes 1ampere = ----
an electric current. Quantitatively, electric current in a I second
conductor across an area held perpendicular to the direction
offlow of charge is defined as the amou n t of charge flowing or
across that area per unit time. Ampere is one basic SI unit. We shall formally define it
If a charge t.Q passes through an area in time t to in chapter 4 in terms of magnetic effect of current.
t + M, then the current I at time t is given by Smaller currents are expressed in following units:
1= lim
t.Q = dQ 1 milliampere = 1 mA = 10-3 A
Ilt -7 a M dt
1 microampere = 1 !iA = 10-6A
(3.1)
, 3.2 PHYSICS-XII
The orders of magnitude of some electric currents
we come across in daily life are as follows:
Current in a domestic appliance ~ 1 A
Formulae Used
Current carried by a lightning ~ 104 A
1. Electric current Charge or I = !i
=
Current in our nerves =-10-6 A = 1 ~A. Time t
ne
4.. Distinguish between conventional and electronic 2. As q = ne, so I = -
t
currents.
3. In case of an electron revolving in a circle of radius
Conventional and electronic currents. By con- r with speed v, period of revolution of the electron is
vention, the direction of motion of positive charges is
R
T = 21tr
taken as the direction of electric current. However, a v
negative charge moving in one direction is equivalent to
Frequency of revolution, v = 2=~
an equal positive charge moving in the opposite T Zrtr
direction, as shown in Fig. 3.1. As the electrons are Current at any point of the orbit is
SI
negatively charged particles, so the direction of electronic I = Charge flowing in 1 revolution
current (i.e., the current constituted by the flow of x No. of revolutions per second
electrons) is opposite to that of the conventional
or I = e v = 3!!.- .
current. 21tr
Units Used
Conventional current Electronic current
~ •• Electric charge is in coulomb (C), time in second
(s), and current in ampere (A)
Constant Used
IT
~I
Charge on an electron, e = 1.6 x 1O-19c.
Fig. 3.1 Flow of negative charge is equivalent to the flow of
positive charge in the opposite direction. Example 1. 1020 electrons, each having a charge of
1.6 x 10-19 C, passfrom a point A towards another point Bin
5. Is electric current a scalar or vector quantity ? 0.1 s. What is the current in ampere? What is its direction?
H
Electric current is a scalar quantity. Although Solution. Here n = 1020, e = 1.6 x 10 -19 C, t = 0.1 s
electric current has both magnitude and direction, yet Current,
it is a scalar quantity. This is because the laws of
ordinary algebra are used to add electric currents and
the laws of vector addition are not applicable to
O
J.ne addition of electric currents. For example, in The direction of current is from B to A.
(Fig. 3.2, two different currents of 3 A and 4 A flowing Example 2. Show that one ampere is equivalent to aflow of
in two mutually perpendicular wires AO and BO meet 1018 elementary charges per second.
6.25 x [CaSE D 92C]
at the junction 0 and then flow along wire Oc.
Solution. Here 1=1 A, t = 1 s, e = 1.6 x 10-19 C
The current in wire OC is 7 A which is the scalar
M
addition of 3 A and 4 A and not 5 A as required by As [=!i=ne
t t
vector addition.
umber of electrons,
li 1x1 u
A n=- = 19 = 6.25 x 10 .
3A e 1.6 x 10-
Example 3. How many electrons pass through a lamp in
90° <'>-0--1--- C one minute, if the current is 300 mA ?
7A
[Himachal 95 ; Punjab 02]
4A
Solution. I = 300 mA = 300 x 10-3 A,
B
t = 1 minute = 60 s, e = 1.6 x 10-19 C
Fig. 3.2 Addition of electric currents is scalar. As [=!i=ne
t t
,CURRENT ELECTRICITY 3.3
.', Number of electrons, Solution. Amount of charge that flows in 10 s
3
. n = It = 300 x 10- x 60 = 1.125 x 1020 = Area under the 1- t graph
e 1.6 x 10-19 = ~ x 5 x 5 + (10 - 5) 5 = 37.5 C
Example 4. How many electrons per second flow through a Example 8. The amount of charge passing through cross-
filament of a 120 V and 60 W electric bulb? Given electric section of a wire is q (t) = at2 + bt + c .
power is the product of voltage and current. (i) Write the dimensional formulae for a, band c.
Solution. Here V = 120 V, P = 60 W, t = 1 s (ii) If the values of a, band c in SI units are 5, 3 and 1
P 60 respectively, find the value of current at t = 5 second.
As P = VI, therefore, I = - = - = 0.5 A
V 120 Solution. (i) Given q (t) = at2 + bt + c
R
Number of electrons,
It 0.5 xl 18
Dimension of a =[t~] = ~; = Ar1
n=- = 19 = 3.125 x 10 .
e 1.6 x 10-
Example 5. In the Bohr model of hydrogen atom, the
Dimension of b = [7] = ~T =A
SI
electron revolves around the nucleus in a circular path of Dimension of c = [q] = AT
radius 5.1 x 10 -11m at afrequency of6.8 x 1015 revolutions
per second. Calculate the equivalent current. (ii) Current, 1= dq = ~ (at2 + bt + c) =2at + b
dt dt
Solution. Here r = 5.1 x 10-11m, At t = 5 s, I= 2 x 5 x 5 + 3 = 53 A.
v =6.8 x 1015 rps, e =1.6 x 10-19 C
Current,
rp roblems For Practice
1= e v = 1.6 x 10-19 x 6.8 x 1015 = 1.088 x 10-3 A. 1. One billion electrons pass from a point P towards
IT
another point Q in 10-3 S . What is the current in
Example 6. In a hydrogen atom, an electron moves in an
ampere? What is its direction?
orbit of radius 5.0 x 10-11 m with a speed of 2.2 x 106 ms-1.
(Ans. 1.6 x 10-7 A, direction of
Find the equivalent current. (Electronic charge = 1.6 x 10-19
current is from Q to P)
coulomb). [Roorkee 84]
2. If 2.25 x 1020
electrons pass through a wire in one
Solution. Here r = 5.0 x 10-11m, minute, find the magnitude of the current flowing
H
v=2.2x106ms-1, e=1.6xlO-19C through the wire. [Punjab 02] (Ans. 0.6 A)
Period of revolution of electron, 3. A solution of sodium chloride discharges
T=2rrr =2rrx5.0x10-
11
s 6.1 x Hy6 N a + ions and 4.6 x 1016Cl" ions in 2 s. Find
the current passing through the solution.
v 2.2 x 106
O
(Ans. 8.56 x 10-3 A)
1 2.2 x 106
Frequency, v = - = -----,., 4. An electric current of 2.0I!A exists in a discharge tube.
T 2rr x 5.0 x 10 11
How much charge flows across a cross-section of
17 the tube in 5 minutes? (Ans. 6.0 x 10-4 C)
2.2x7x10 =7x1015s-1
2x22x5 5. In a hydrogen atom, the electron makes about
M
Current, 1= e v = 1.6 x 10-19 x 7 x 1015 0.6 x Hy6 revolutions per second around the nucleus.
Determine the average current at any point on the
= 1.12 x 10-3 A.
orbit of the electron. (Ans. 0.96 mA)
Example 7. Figure 3.3 shows a plot of current I through 6. An electron moves in a circular orbit of radius 10 em
the cross-section of a wire over a time interval of10 s. Find with a constant speed of 4.0 x 106 ms-1. Determine
the amount of I(A) the electric current at a point on the orbit.
charge that flows (Ans. 1.02 x 10-12 A)
through the wire 5
during this time 7. In a hydrogen discharge tube, the number of
protons drifting across a cross-section per second is
period.
[CBSE00 lSC]
1.1 x 1018, while the number of electrons drifting in
the opposite direction across another cross-section
5 10 t (s)
is 3.1 x 1018 per second. Find the current flowing in
Fig. 3.3 the tube. (Ans. 0.672 A)
, 3.4 PHYSICS-XII
HINTS A steady flow of electric current in a conductor is
9 19 maintained in a similar way. As shown in Fig. 3.5,
1. [= ne = 10 x 1.6 x 10- =1.6 X 10-7 A
positive charge flows spontaneously in a conductor
t 10-3
from higher potential (A) to lower potential (B) i.e., in
ne 2.25 x 1020 x 1.6 x 10-19
2. [= - = = 0.6 A the direction of the electric field. To maintain steady
t 60 current through the conductor, some external device
(n+ + n-)e must do work at a steady rate to take positive charge
3. [= [cations + [anions = t
from lower potential (B) to the higher potential (A).
(6.1x1016 +4.6x1016)x1.6 x 10-19 -3 Such a device is the source of electromotive force (emf)
-'---------'------ = 8.56 x10 A which may be an electrochemical cell or an electric-
2
generator. A source of emf transfers positive charge
R
4. q = It = 2.0 x10-6 x5 x60 = 6.0 x10-4 C. form lower potential to higher potential i.e., in the
5. [= ve = 0.6 X 1016 X 1.6 x 10-19 opposite direction of the electric field. Clearly, a charge
= 0.96 X 10-3 A = 0.96 mA flow circuit is analogous to the water flow circuit.
6
6. T = 21tr = 21t x O.!O s :. V = ~ = 4 x 10 s-1
SI
V 4 x 10 T 21t X 0.10 A-:~ Source of emf
19 B - (Charge pump)
l __ 4x106 x 1.6 x 10- -102 10-12A
=ve - -. X • R
21t xO.10
(n + ne) e
7. [= l + I = --,-P--- Fig. 3.5 A closed charge flow circuit.
P" t
(1.1 X 1018 + 3.1 X 1018) x 1.6 X 10-19 3.4 ELECTROMOTIVE FORCE : EMF
= = 0.672A
1
7. Define emf of a battery. Is it really aforce? When is
IT
the emf of a battery equal to the potential difference
3.3 MAINTENANCE OF STEADY CURRENT
between its terminals? Define emf of 1 volt.
IN A CIRCUIT
Electromotive force. A battery is a device which
6. With the help of a mechanical analogy, explain how maintains a potential difference between its two ter-
f+-
~ theflow of electric current is maintained in an electric circuit. minals A and B.
---B ·~-------------I
Maintenance of steady current in an electric circuit.
H
The flow of electric current in a circuit is analogous to
F, •
the flow of water in a pipe. As shown in Fig. 3.4,
suppose we wish to maintain a steady flow of water in : A Fe B I
a horizontal pipe from A to B. As pressure at A is
higher than that at B, so water flows spontaneously Fig. 3.6 A schematic diagram of a battery.
O
from the upper tank to the lower tank. To maintain a
steady flow, a water pump must do work at a steady Figure 3.6 shows a schematic diagram of a battery.
rate to pump water back from the lower tank to the Due to certain chemical reactions, a force (of non-
upper tank. Obviously, the water pump makes water electrostatic origin) is exerted on the charges of the
flow from lower to higher pressure. It helps to maintain electrolyte. This force drives positive charges towards
the pressure difference between A and B. terminal A and negative charges towards terminal B.
M
-+
Suppose the force on a positive charge q is F". As the
-11\
charges build up on the two terminals A and B, a
'n\
potential difference is set up between them. An electric
I
-+
field E is set up in the electrolyte from A to B.This field
-+ -+
h
Water exerts a force Fe = q E on the charge q, in the opposite
pump -+
direction of Fn. In the steady state, the charges stop
l~A======~ ..'
accumulating further and F" = Fe .
The work done by the non-electrostatic force during
the displacement of a charge q from B to A is
W=Fn d
Fig. 3.4 A closed water flow circuit. where d is the distance between the terminals A and B.
Cu RRENT ELECTRICITY
R
SI
3.1 CURRENT ELECTRICITY If the current is steady i.e., the rate of flow of charge
1. What is current electricity ? does not change with time, then
Current electricity. In chapters 1 and 2, we studied 1= Q
IT
the phenomena associated with the electric charges at t
rest. The physics of charges at rest is called . Electric charge
electrostatics or static electricity. We shall now study or EIectnc current = ------"'-
Time
the motion or dynamics of charges. As the term current
implies some sort of motion, so the motion of electric where Q is the charge that flows across the given area
charges constitutes an electric current. in time t.
H
The study of electric charges in motion is called current Lightning, which is the flow of electric charge
electrici ty. between two clouds or from a cloud to the earth, is an
example of a transient current (a current of short
3.2 ELECTRIC CURRENT duration). But the charges flow in a steady manner in
devices like a torch, cell-driven clock, transistor radios,
O
2. Define electric current. hearing aids, etc.
El~tric current. If two bodies charged to different 3. Give the 51 unit of current.
poten .als are connected together by means of a
SI unit of current is ampere. If one coulomb of charge
conduc ing wire, charges begin to flow from one body
crosses an area in one second, then the current through that
to another. The charges continue to flow till the
area is one ampere (A).
M
potentials of the two bodies become equaL
1coulomb
Theflow of electric charges through a conductor constitutes 1ampere = ----
an electric current. Quantitatively, electric current in a I second
conductor across an area held perpendicular to the direction
offlow of charge is defined as the amou n t of charge flowing or
across that area per unit time. Ampere is one basic SI unit. We shall formally define it
If a charge t.Q passes through an area in time t to in chapter 4 in terms of magnetic effect of current.
t + M, then the current I at time t is given by Smaller currents are expressed in following units:
1= lim
t.Q = dQ 1 milliampere = 1 mA = 10-3 A
Ilt -7 a M dt
1 microampere = 1 !iA = 10-6A
(3.1)
, 3.2 PHYSICS-XII
The orders of magnitude of some electric currents
we come across in daily life are as follows:
Current in a domestic appliance ~ 1 A
Formulae Used
Current carried by a lightning ~ 104 A
1. Electric current Charge or I = !i
=
Current in our nerves =-10-6 A = 1 ~A. Time t
ne
4.. Distinguish between conventional and electronic 2. As q = ne, so I = -
t
currents.
3. In case of an electron revolving in a circle of radius
Conventional and electronic currents. By con- r with speed v, period of revolution of the electron is
vention, the direction of motion of positive charges is
R
T = 21tr
taken as the direction of electric current. However, a v
negative charge moving in one direction is equivalent to
Frequency of revolution, v = 2=~
an equal positive charge moving in the opposite T Zrtr
direction, as shown in Fig. 3.1. As the electrons are Current at any point of the orbit is
SI
negatively charged particles, so the direction of electronic I = Charge flowing in 1 revolution
current (i.e., the current constituted by the flow of x No. of revolutions per second
electrons) is opposite to that of the conventional
or I = e v = 3!!.- .
current. 21tr
Units Used
Conventional current Electronic current
~ •• Electric charge is in coulomb (C), time in second
(s), and current in ampere (A)
Constant Used
IT
~I
Charge on an electron, e = 1.6 x 1O-19c.
Fig. 3.1 Flow of negative charge is equivalent to the flow of
positive charge in the opposite direction. Example 1. 1020 electrons, each having a charge of
1.6 x 10-19 C, passfrom a point A towards another point Bin
5. Is electric current a scalar or vector quantity ? 0.1 s. What is the current in ampere? What is its direction?
H
Electric current is a scalar quantity. Although Solution. Here n = 1020, e = 1.6 x 10 -19 C, t = 0.1 s
electric current has both magnitude and direction, yet Current,
it is a scalar quantity. This is because the laws of
ordinary algebra are used to add electric currents and
the laws of vector addition are not applicable to
O
J.ne addition of electric currents. For example, in The direction of current is from B to A.
(Fig. 3.2, two different currents of 3 A and 4 A flowing Example 2. Show that one ampere is equivalent to aflow of
in two mutually perpendicular wires AO and BO meet 1018 elementary charges per second.
6.25 x [CaSE D 92C]
at the junction 0 and then flow along wire Oc.
Solution. Here 1=1 A, t = 1 s, e = 1.6 x 10-19 C
The current in wire OC is 7 A which is the scalar
M
addition of 3 A and 4 A and not 5 A as required by As [=!i=ne
t t
vector addition.
umber of electrons,
li 1x1 u
A n=- = 19 = 6.25 x 10 .
3A e 1.6 x 10-
Example 3. How many electrons pass through a lamp in
90° <'>-0--1--- C one minute, if the current is 300 mA ?
7A
[Himachal 95 ; Punjab 02]
4A
Solution. I = 300 mA = 300 x 10-3 A,
B
t = 1 minute = 60 s, e = 1.6 x 10-19 C
Fig. 3.2 Addition of electric currents is scalar. As [=!i=ne
t t
,CURRENT ELECTRICITY 3.3
.', Number of electrons, Solution. Amount of charge that flows in 10 s
3
. n = It = 300 x 10- x 60 = 1.125 x 1020 = Area under the 1- t graph
e 1.6 x 10-19 = ~ x 5 x 5 + (10 - 5) 5 = 37.5 C
Example 4. How many electrons per second flow through a Example 8. The amount of charge passing through cross-
filament of a 120 V and 60 W electric bulb? Given electric section of a wire is q (t) = at2 + bt + c .
power is the product of voltage and current. (i) Write the dimensional formulae for a, band c.
Solution. Here V = 120 V, P = 60 W, t = 1 s (ii) If the values of a, band c in SI units are 5, 3 and 1
P 60 respectively, find the value of current at t = 5 second.
As P = VI, therefore, I = - = - = 0.5 A
V 120 Solution. (i) Given q (t) = at2 + bt + c
R
Number of electrons,
It 0.5 xl 18
Dimension of a =[t~] = ~; = Ar1
n=- = 19 = 3.125 x 10 .
e 1.6 x 10-
Example 5. In the Bohr model of hydrogen atom, the
Dimension of b = [7] = ~T =A
SI
electron revolves around the nucleus in a circular path of Dimension of c = [q] = AT
radius 5.1 x 10 -11m at afrequency of6.8 x 1015 revolutions
per second. Calculate the equivalent current. (ii) Current, 1= dq = ~ (at2 + bt + c) =2at + b
dt dt
Solution. Here r = 5.1 x 10-11m, At t = 5 s, I= 2 x 5 x 5 + 3 = 53 A.
v =6.8 x 1015 rps, e =1.6 x 10-19 C
Current,
rp roblems For Practice
1= e v = 1.6 x 10-19 x 6.8 x 1015 = 1.088 x 10-3 A. 1. One billion electrons pass from a point P towards
IT
another point Q in 10-3 S . What is the current in
Example 6. In a hydrogen atom, an electron moves in an
ampere? What is its direction?
orbit of radius 5.0 x 10-11 m with a speed of 2.2 x 106 ms-1.
(Ans. 1.6 x 10-7 A, direction of
Find the equivalent current. (Electronic charge = 1.6 x 10-19
current is from Q to P)
coulomb). [Roorkee 84]
2. If 2.25 x 1020
electrons pass through a wire in one
Solution. Here r = 5.0 x 10-11m, minute, find the magnitude of the current flowing
H
v=2.2x106ms-1, e=1.6xlO-19C through the wire. [Punjab 02] (Ans. 0.6 A)
Period of revolution of electron, 3. A solution of sodium chloride discharges
T=2rrr =2rrx5.0x10-
11
s 6.1 x Hy6 N a + ions and 4.6 x 1016Cl" ions in 2 s. Find
the current passing through the solution.
v 2.2 x 106
O
(Ans. 8.56 x 10-3 A)
1 2.2 x 106
Frequency, v = - = -----,., 4. An electric current of 2.0I!A exists in a discharge tube.
T 2rr x 5.0 x 10 11
How much charge flows across a cross-section of
17 the tube in 5 minutes? (Ans. 6.0 x 10-4 C)
2.2x7x10 =7x1015s-1
2x22x5 5. In a hydrogen atom, the electron makes about
M
Current, 1= e v = 1.6 x 10-19 x 7 x 1015 0.6 x Hy6 revolutions per second around the nucleus.
Determine the average current at any point on the
= 1.12 x 10-3 A.
orbit of the electron. (Ans. 0.96 mA)
Example 7. Figure 3.3 shows a plot of current I through 6. An electron moves in a circular orbit of radius 10 em
the cross-section of a wire over a time interval of10 s. Find with a constant speed of 4.0 x 106 ms-1. Determine
the amount of I(A) the electric current at a point on the orbit.
charge that flows (Ans. 1.02 x 10-12 A)
through the wire 5
during this time 7. In a hydrogen discharge tube, the number of
protons drifting across a cross-section per second is
period.
[CBSE00 lSC]
1.1 x 1018, while the number of electrons drifting in
the opposite direction across another cross-section
5 10 t (s)
is 3.1 x 1018 per second. Find the current flowing in
Fig. 3.3 the tube. (Ans. 0.672 A)
, 3.4 PHYSICS-XII
HINTS A steady flow of electric current in a conductor is
9 19 maintained in a similar way. As shown in Fig. 3.5,
1. [= ne = 10 x 1.6 x 10- =1.6 X 10-7 A
positive charge flows spontaneously in a conductor
t 10-3
from higher potential (A) to lower potential (B) i.e., in
ne 2.25 x 1020 x 1.6 x 10-19
2. [= - = = 0.6 A the direction of the electric field. To maintain steady
t 60 current through the conductor, some external device
(n+ + n-)e must do work at a steady rate to take positive charge
3. [= [cations + [anions = t
from lower potential (B) to the higher potential (A).
(6.1x1016 +4.6x1016)x1.6 x 10-19 -3 Such a device is the source of electromotive force (emf)
-'---------'------ = 8.56 x10 A which may be an electrochemical cell or an electric-
2
generator. A source of emf transfers positive charge
R
4. q = It = 2.0 x10-6 x5 x60 = 6.0 x10-4 C. form lower potential to higher potential i.e., in the
5. [= ve = 0.6 X 1016 X 1.6 x 10-19 opposite direction of the electric field. Clearly, a charge
= 0.96 X 10-3 A = 0.96 mA flow circuit is analogous to the water flow circuit.
6
6. T = 21tr = 21t x O.!O s :. V = ~ = 4 x 10 s-1
SI
V 4 x 10 T 21t X 0.10 A-:~ Source of emf
19 B - (Charge pump)
l __ 4x106 x 1.6 x 10- -102 10-12A
=ve - -. X • R
21t xO.10
(n + ne) e
7. [= l + I = --,-P--- Fig. 3.5 A closed charge flow circuit.
P" t
(1.1 X 1018 + 3.1 X 1018) x 1.6 X 10-19 3.4 ELECTROMOTIVE FORCE : EMF
= = 0.672A
1
7. Define emf of a battery. Is it really aforce? When is
IT
the emf of a battery equal to the potential difference
3.3 MAINTENANCE OF STEADY CURRENT
between its terminals? Define emf of 1 volt.
IN A CIRCUIT
Electromotive force. A battery is a device which
6. With the help of a mechanical analogy, explain how maintains a potential difference between its two ter-
f+-
~ theflow of electric current is maintained in an electric circuit. minals A and B.
---B ·~-------------I
Maintenance of steady current in an electric circuit.
H
The flow of electric current in a circuit is analogous to
F, •
the flow of water in a pipe. As shown in Fig. 3.4,
suppose we wish to maintain a steady flow of water in : A Fe B I
a horizontal pipe from A to B. As pressure at A is
higher than that at B, so water flows spontaneously Fig. 3.6 A schematic diagram of a battery.
O
from the upper tank to the lower tank. To maintain a
steady flow, a water pump must do work at a steady Figure 3.6 shows a schematic diagram of a battery.
rate to pump water back from the lower tank to the Due to certain chemical reactions, a force (of non-
upper tank. Obviously, the water pump makes water electrostatic origin) is exerted on the charges of the
flow from lower to higher pressure. It helps to maintain electrolyte. This force drives positive charges towards
the pressure difference between A and B. terminal A and negative charges towards terminal B.
M
-+
Suppose the force on a positive charge q is F". As the
-11\
charges build up on the two terminals A and B, a
'n\
potential difference is set up between them. An electric
I
-+
field E is set up in the electrolyte from A to B.This field
-+ -+
h
Water exerts a force Fe = q E on the charge q, in the opposite
pump -+
direction of Fn. In the steady state, the charges stop
l~A======~ ..'
accumulating further and F" = Fe .
The work done by the non-electrostatic force during
the displacement of a charge q from B to A is
W=Fn d
Fig. 3.4 A closed water flow circuit. where d is the distance between the terminals A and B.
