Mod-3 Epotential New

CHAPTER 3 •

THE ELECTRIC POTENTIAL o 1. Introduction o 2. Equi-potential surfaces o 3. Calculating the Electric Potential  Example problem 3.1 Potential difference between two points  Example: Problem 3.2 Potential due to a Point Charge  Example: Problem 3.3 Potential due to a Point Charge  Example: Problem 3.3 Potential due to a Sphere of charge 4. Electric Potential due to a system of point charges  Example: Problem 4.1  5. Electric Potential due to continuous distribution of charges  Example: Problem 5.1 Electric potential due to a line of charge  Example: Problem 5.2 Electric potential due to a ring of charge  Example: Problem 5.3 Electric potential due to a disk of charge o 6. The Gradient of the Electric Potential  Example: Problem 5.1  Example: Problem 5.2 o 7. The Potential and Field of a Dipole o 8. Energy of a system of charges

THE ELECTRIC POTENTIAL 1. Introduction The electric force is a conservative force. Whenever a system of particles interacts with each other by means of electric force, that is conservative in nature, we can assign potential energy to the system. (A force is conservative when the work it does on a particle depends only on the initial and final position of the particle, and not on the path followed). The introduction of the potential energy is useful since it allows us to apply conservation of mechanical energy that simplifies the solution of a large number of problems. The potential energy U associated with a conservative force F is defined in the following manner. The change in potential energy of a particle which moves from position i to f in a force field F is given by f

U f − U i = − Wif = − ∫ F • d l

…….. …….

(1)

i

where Wif is the work done by the force during the move from reference position i to f . The path integral is done along any convenient path connecting i and f. Since the force F is conservative, the integral in eq.(1) will not depend on the path chosen. If the work W if is positive (force and displacement pointing in the same direction) the potential energy at f will be smaller than the potential energy at i. If energy is conserved, a decrease in the potential energy will result in an increase of the kinetic energy. If the work W is negative (force and displacement pointing in opposite directions) the potential energy at f will be larger than the potential energy at i. If energy is conserved, an increase in the potential energy will result in an decrease of the kinetic energy. In electrostatic problems the potential energy for a charged particle is determined at a point of an electric field. Here the reference point i is usually chosen at an infinite distance, and the potential energy at this reference point is taken to be equal to zero. With Ui = 0 the f

potential energy at the point f is:

U f = − W∞ = − ∫ F • d l

……

…….

∞

…… (2) The potential energy of the particle at position f is equal to the negative of the work done by the force the to bring it from infinity.

To describe the potential energy associated with a charge distribution the concept of the electric potential (V) is introduced. The electric potential V at a given position is defined as the potential energy of a test particle divided by the charge q of this object:

Vf =

Uf

1 =− q

q

f

∫ F • d l …….

…….. …… (3)

∞

In the above calculation we have assumed that the reference point i is at infinity, and that the electric potential at that point is equal to 0. Since the force per unit charge( F/q) is the electric field (see Chapter 1), eq. (3) can be rewritten as f

Vf = −

∫ E • dl

…….

……..

……. (4)

∞

i.e. the electric potential at a point of electric field is equal to the negative of the work done by the electric field while bringing a unit positive test charge from infinity to that point. The unit of electric potential is the volt (V), and 1 V = 1 J/C = 1 Nm/C. Equation (4) shows that as the unit of the electric field we can also use V/m. The potential difference between two point i and f in principal can be written as

V f − Vi = −

Wif q

f

=−

Electron volt unit of energy

∫

E • dl

........ ……

…… (5)

∞

A common used unit for the energy of a particle is the electron-volt (eV) which is defined as the change in kinetic energy of an electron that travels over a potential difference of 1 V. The electron-volt can be related to the Joule via eq.(3). Equation (3) shows that the change in energy of an electron when it crosses over a 1 V potential difference is equal to 1.6 . 10-19 J and we thus conclude that 1 eV = 1.6 x10-19 J

2. Equi-potential surfaces Equipotential surfaces are defined as surfaces on which each point has the same electric potential. The component of the electric field parallel to this surface must be zero since the change in the potential between all points on this surface is equal to zero. This implies that the direction of the electric field is perpendicular to the equipotential surfaces. Figure below shows a family of equipotential surfaces associated with an electric field due to some charge distribution. The work done by the electric field on a test charge q as it moves from point I to point II is zero as the end points are situated on the same equipotential surface. Again the work done I II by the electric field on the test charge as it moves from III IV point III to point IV is also zero because of the same reason as stated above. However the work done by V VII electric field to transfer a test charge from point VII to VI VIII point VIII is equal to the work done by electric field to transfer the charge from V to VI as the end points of these two paths are situated on the same set of equipotential surfaces. In the following figures the electric fields are represented by solid lines where as the corresponding equi-potential surfaces are represented by dotted lines. Field lines

Equi-potential surfaces Fig 2a : Equi-potential surfaces in uniform field

Fig 2b : Equi-potential surfaces in a field of point charge

3. Calculating the Electric Potential Example problem 3.1 Potential difference between two points Figure below shows two points i and f in an electric field E. the points lie on the same field line and are separated by d. calculate the potential difference V f – Vi (a) by moving a test charge along the path parallel to the field direction. (b) by moving a test charge along the path shown in figure (b) i c

i

d

dl 450 d

dl

f

f Figure (3.1a)

Figure (3.1b)

(a) For the path shown in figure (a)

f

E. dl = E dl cos 00 = E dl

Then the integral

By definition potential difference between I and f is

−

f

f

∫ E • d l = − ∫ E dl = − E ∫ dl = − Ed i

i

∫ E . d l is divided into two i

part i.e. f

c

f

i

i

c

∫ E • dl = ∫ E • dl + ∫ E • dl

Vf – Vi = f

(b) For path shown in figre b the charge is farst moved along ic and then along cf path.

i

For path ‘if’ E. dl = E dl cos 900 = 0 and for path ‘fc’ E. dl = E dl cos 450 = E dl/√2 then the path integral is equal to f

∫ E • dl = c

Hence

E

f

∫ dl = 2 c

E 2

2d =Ed

Vf – Vi = - Ed

That is, the work required to transfer a charge from one point to another does not depend on the path followed.

Example problem 3.2 Potential due to a point charge (q) A charge q0 is moved from i to f in the vicinity of charge q (see Figure 3.2) The electric potential at f can be determined using eq. (4) and evaluating the integral along the path shown in Figure 3.1. Changing the variable from l to r′ equation (4) can therefore be rewritten as r

Vr = − ∫ E dr ′

……..

……..

………. (6)

∞

Here the initial point has been taken at infinity (∞) where potential is considered to be zero. The magnitude of electric field at the site of the test charge is

E=

1 q 4πε o r ′ 2

i

Substituting this value in the previous equation we get

r∞

r

Vr =

−

1 q ∫∞ 4πε o r ′ 2 dr ′

q =− 4πε o

r

∞

f

dr

∫r

2

qo

r q Figure 3.2 : Path followed by charge qo between i

=−

r

 1  − r  ∞

q 4πε o

=

1 q 4πε o r

…..

……

….. ….. (7)

If the charge q is positive, the potential increases with a decreasing distance r. The electric field points away from a positive charge, and we conclude that the electric field points from regions with a high electric potential towards regions with a low electric potential.

Example problem 3.3 Potential due to a sphere of charge (i) determine electric potential at a point outside the sphere (ii) determine electric potential at a point inside the sphere Q charge is uniformly distributed over a sphere of radius R. We know that the electric field outside and inside the sphere of charge are given by

1 Q 4π ∈o r 2

……….

……..

……… (4a)

  Q r E A =  3   4π ∈o R 

……….

……….

………. (4b)

EB = and

+ + + + + + R + + + + + + + + + + + + + + + + + + + + ++

Electric potential at any point outside the sphere can be determined just by evaluating the following integral r

Vr = − ∫ E dr ′ ∞

Here the initial point has been taken at infinity (∞) where potential is considered to be zero. Substituting the expression of E from eq (4a) we get r

Vr =

Q dr ′ 2 ′ 4 πε o r ∞

Q = − 4πε o =

1

−∫

r

dr ′

∫ r′

2

∞

1 Q 4πε o r

……….

……..

……

(5)

The above expression gives the electric potential at points outside the sphere of charge.The potential at its surface be found by putting r = R in the above expression, i.e.

VR =

1 Q 4πε o R

………

……..

……..

(6)

Now the potential difference between a point inside the sphere and a point on the surface of the sphere is given by r

Vr - V R =

− ∫ E dr ′

……….

……….

……… (7)

R

Using the expression of E inside the sphere as given in equation ( 5 ) the above equation can be written as r

Vr - V R =

−

Q r ′ dr ′ 4π ∈o R 3 ∫R

r

 r ′2  Q or, Vr - VR = −   4π ∈o R 3  2  R or, Vr - VR =

Putting

VR =

−

(

Q r 2 − R2 3 8π ∈o R

)

1 Q the above equation can be re written as 4πε o R

Vr = =

(

Q Q − r2 − R2 3 4π ∈o R 8π ∈o R

(

Q 3R 2 − r 2 3 8π ∈o R

)

)

…… ………. ……. (8) The above equation gives the electric potential at points inside a uniformly charged sphere a distance of r from its center. Potential energy of a test charge in the electric field From the definition of the electric potential in terms of the potential energy (eq.3) it is clear that the potential energy of a charge qo under the influence of the electric field generated by charge q is given by

U r = q o Vr =

1 q qo …… 4πε o r

….

…..

(8)

Example: Problem 3.3 An alpha particle with a kinetic energy of 1.7 x 10-12 J is shot directly towards a platinum nucleus from a very large distance. What will be the distance of closest approach? The electric charge of the alpha particle is 2e and that of the platinum nucleus is 78e. Treat the alpha particle and the nucleus as spherical charge distributions and disregard the motion of the nucleus.

Solution: The initial mechanical energy (E = U + K) is equal to the kinetic energy of the alpha particle Ei = Ki = 1.7 x 10-12 J …… ….. ……. (9) Due to the electric repulsion between the alpha particle and the platinum nucleus, the alpha particle will slow down. At the distance of closest approach the velocity of the alpha particle is zero, and thus its kinetic energy is equal to zero. The total mechanical energy at this point is equal to the potential energy of the system Ef =

Uf =

1 q qo …… 4πε o d

…….

…..

….. (10)

Where qo is the charge of the alpha particle, q is the charge of the platinum nucleus, and d is the distance of closest approach. Applying conservation of mechanical energy we obtain Ef =

Uf =

1 q qo 4πε o d

= Ei = 1.7 x 10-12 J ….

…..

The distance of closest approach can be obtained from eq. (10)

d=

q qo 2 e x 78e 1 = 9 x 10 9 = 2.1 x 10 −14 m −12 −12 4πε o 1.7 x 10 1.7 x 10

4 Electric Potential due to a system of point charges

….. (11)

According to superposition principal the electric potential of a system of point charges is equal to the sum of the potentials created at a given point by each of the charges. Consequently using formula (7) we can write the following formula for potential due to a system of point charges

V = ∑Vi =

1 4πε o

n

qi

i =1

i

∑r

…..

(11)

Four point charges are placed at the corners of a square as shown in figure below. Calculate the electric potential at its center. [ here q1 = 12 nC, q2 = -24 nC, q3= 31 nC , q4 = 17 nC and d = 1.3 m]

q1

d

q2

Distance of the corners from the center of the square is r = 1.3/√2 m Now using equation (11) the potential at the center is obtained as

d

o

d

q3  q1 q 2 q3 q 4  + +  +  r r r r4  1 2 3  12 x 10 −9 − 24 x 10 −9 31 x 10 −9 7 x 10 −9  = 9 x 10 9  + + +  1.3 / 2 1.3 / 2 1.3 / 2   1.3 / 2 = 254.56V

d

q4

i

….

…..

Here, ri is the radial distance of the point from the charge qi.

Example problem 4.1

1 V = ∑ Vi = 4πε o i

qi 1 = ∑ 4πε o i =1 ri 4

(

)

5. Electric Potential due to continuous distribution of charges Electric Potential due volume charges Let us assume that the charges are distributed over finite region of space and that potential is normalized to zero at infinity. Denoting volume charge density by ρ we obtain the following expression for potential instead of (11)

V =

1 4πε o

∫

ρ dV r

….

…..

…..

(12)

where dV is the volume element over which integration is performed. Electric Potential due to surface charges If a charge is located on a surface , the charge distribution is characterized by surface charge density σ. On the area element dA there is charge σ dA, and hence the potential at a given point is given by

V =

1 σ dA ∫ 4πε o S r

….

…..

…..

(13)

where r is the distance between the area element dA and the point at which potential is calculated.

Example problem 5.1 : Electric potential due to a line of charge A total charge Q is distributed uniformly along a straight rod of length L. Find the potential P at point P at a distance d from the midpoint of the rod (Figure 2). The potential at P due to a small segment of

d

the rod, with length dx and charge dQ, located at

-L/2

dx x

+L/2

the position indicated in Figure 2 is given by

dV =

1 4πε o

dQ

……

x + d2 2

(14 )

The charge dQ of the segment is

dQ =

Q dx L

…….

…….

………

(15 )

Combining equations (14 ) and (15 ) we obtain the following expression for dV:

dV =

1 Q 4πε o L

dx x + d2 2

…..

……

…..

(16 )

the total potential at P can be obtained by summing over all small segments. This is equivalent to integrating eq.() between x = - L/2 and x = L/2.

1 Q V = 4πε o L =

L/2

∫

−L / 2

[

dx x + d2 2

1 Q ln x + x 2 + d 2 4πε o L

]

L/2 −L/2

L  L2 + + d2   4 2  1 Q = ln  4πε o L  L  L2 − + + d2   4  2 

……

…. …..

(17)

Example: Problem 5.2 Electric potential due to a ring of charge A thin ring of insulating material has a radius R. An amount of charge Q is uniformly distribute over it . Find the potential as a function of the distance on the axis of the ring. Solution : We define the x-axis to coincide with the axis of the ring (Figure 5.2). The first step in the calculation of the total electric potential at point P due to the annulus is to calculate the electric potential at P due to a small segment of the ring . Consider a segment with length dl as shown in Figure5.2. The electric potential dV at P generated by this ring is given by

dV =

1 4πε o

dQ x2 + R2

…..

…..

…. (18)

where dQ is charge on dl length of the ring. If ρ is the linear charge density on the ring

Q ρ= 2πR

dl

…..

…..

…..

….. (19)

R O

x

Figure : 5.2

P

using eq. (25.33) the charge dQ of the ring can be calculates

Q dl …. 2πR

dQ = ρ dl =

….

…. (20)

Substituting eq.(20) into eq.(18) we obtain

dV =

1 Q 4πε o 2πR

dl

….

x + R2 2

…..

…..

(21)

The total electric potential can be obtained by integrating eq.(21) over the whole ring:

V =

or

1 4πε o

V =

Q

∫ 2πR

dl x +R

1 Q 4πε o 2πR

2

=

2

1 x +R 2

2

1 Q 4πε o 2πR

( 2πR ) =

1 4πε o

1 x + R2 2

∫ dl

Q

…. ……. ……(21)

x + R2 2

Example Problem 5.3 Electric potential due to a disk of charge A disk made of insulator has a radius R and an amount of electric charge Q is uniformly distributed over it. Find the potential as a function of the distance on the axis of the disk. Solution : We define the x-axis to coincide with the axis of the ring (Figure 5.3). The first step in the calculation of the total electric potential at point P due to the disk is to calculate the electric potential at P due to a small segment of the disk. Consider a ring with radius r and width dr as shown in Figure 5.3. The electric potential dV at P generated by this ring is

dV =

given by …..

1 4πε o

dQ x + r2 2

………. …….

(22)

where dQ is the charge on the ring. The charge density σ of the disk is equal to

σ =

Q Q = A πR 2

…..

…..

…..

(23)

…..

(24)

z =

r P R Figure 5.3. Problem 5.3. Using eq. (23) the charge dQ of the ring can be calculated as

Q 2Q dQ = σ dA = 2 2πr dr = 2 r dr ….. πR R

…..

Substituting eq.(24) into eq.(22) we obtain

dV =

r dr

1 2Q 4πε o R 2

….

x2 + r 2

….

….

(25)

The total electric potential can be obtained by integrating eq.(25) over the whole disk

1 2Q V = 4πε o R 2

R

∫ 0

r dr x2 + r 2

1 1 2Q = 4πε o 4πε o R 2 1 Q = 2πε o R 2

[ (

x2 + r 2

]

R o

x + R −x 2

2

)

…..

….

(26)

(a) when P is at a large distance from the center of the disk (x>>R) then the equation can be written as

V=

(

1 Q  2 x + R2 2  2πε o R 

)

1

2

− x  

1  1 Qx   R2  2    1 + − 1 or V = 2  2   2πε o R   x    2   1 Qx   1R  1 + = + ....... −1 ….. (27) 2  2 2πε o R   2 x  

Neglecting the higher order terms of the binomial ` series we can write

V=

1 Qx  1 R 2  1 Q   = 2  2  2πε o R  2 x  4πε o x

( point charge)

(b) at the center of the disk at the center of the disk x =0. so the electric potential is given by

V=

(

1 Q  2 x + R2 2  2πε o R 

)

1

2

1 Q ( R ) = 1 Q ….. − x  = 2  2πε o R 2πε o R

…..

..

(28)

6. The Gradient of the Electric Potential The electric potential V is related to the electric field E. If the electric field E is known, the electric potential V can be obtained using eq.(4), and vice-versa. In this section we will discuss how the electric field E can be obtained if the electric potential is known. Let us consider two equipotential surface differing in potential by dV. E is perpendicular to the surfaces at any point. . a test charge q0 is moved from point 1 to point 2 situated on surface 1 and 2 respectively along path dl the work done by the electric field during the move is then dW = qo E. dl = qo E dl cosθ again in terms of potential difference dV the work done is dW = -qo dV so that we can write

dW = qo E dl cosθ = -qo dV or,

(E cosθ) dl = - dV

or

E cosθ =

−

∂V ∂l

or

El =

−

∂V ∂l

…..

…….

…..

(29)

here El the component of E along l axis. If the direction of the displacement is chosen to coincide with the x-axis, eq.(29) becomes

Ex = −

∂V ∂x

…….

…….

…..

(30)

Similarly the y an z component of e can by given by

and

Ey = −

∂V ∂y

Ez = −

∂V ∂z

The total electric field E can be obtained from the electric potential V by combining equations (29), (30), and (31):

……

…….. …….

(32)

Equation (32) is usually written in the following form

(25.23)

…….. ……… ……… (33)

Example: Problem 6.1 In some region of space, the electric potential is the following function of x, y, and z:

where the potential is measured in volts and the distances in meters. Find the electric field at the points x = 2 m, y = 2 m. The x, y and z components of the electric field E can be obtained from the gradient of the potential V

and

Evaluating equations (a), (b), and (c) at x = 2 m and y = 2 m gives

Thus

7. The Potential and Field of a Dipole Figure 7 shows an electric dipole located along the z-axis. It consists of two charges + Q and - Q, separated by a distance L. The electric potential at point P can be found by summing the potentials generated by each of the two charges:

…….. ……

….. (34)

Figure 7.1. The electric dipole. If the point P is far away from the dipole (r >> L) we can make the approximation that r1 and r2 are parallel. In this case .

and

…….. …..

…..

(35)

……….

……. …….. (36)

The electric potential at P can now be rewritten as

…..

…..

…….

(37)

Where p is the dipole moment of the charge distribution.

8. Electric potential energy of a system of point charges The electric potential energy of a system of point charges is equal to the negative of work done by electric field to assemble the system of charges from infinite separation. Let us consider two charges q1 and q2 are initially separated by a distance of ri. If their separation is changed to rf the work done by electric field is f

Wif =

∫ F • dl i

here electric force between the charges is

F =

1 q1 q 2 rˆ 4πε o r 2

and

d l = dr rˆ

where dr is the element f radius vector. Then the work done by electric field can be evaluated as

Wif

1 = 4πε o

f

∫ i

q1 q 2 q q dr = 1 2 2 4πε o r

 −

r

q q 1 f = − 1 2  r  ri 4πε o

1 1   −   r f ri 

By definition the potential energy at final configuration is

U = − W12

= =

q1 q 2 1 4πε o r12

(where rf = r12 and ri = ∞)

If a third charge is brought in the scenario then the potential energy is calculated by bringing the charges one by one and calculating the work done by the electric force and taking summation in the following manner: First bring charge q1 which requires no work as electric field is zero. Then bring charge q2 in the field of q1 . The work done by the electric force is

W12 = −

q1 q 2 1 4πε o r12

now bring charge q3 in the field of q1 and q2 . The work done by the fields of these charges are

W13 = −

q1 q3 1 4πε o r13

and

W23 = −

q2 q3 1 respectively 4πε o r23

Then the potential energy of the configuration f these three charges is

U = − (W12 + W13 + W23

)

=

1 4πε o

 q1 q 2 q q q q  + 1 3 + 2 3 r13 r23  r12

  

If there are n charges the general expression for potential energy of the system of charges is

U =

1 2

 1   4πε o 

n

qi q j    r j =1 ij  n

∑∑ i =1

; ( j ≠ i)

here the factor ½ has been included to compensate for each pair of charges counted twice.

Problem:

Exercise from Electric Potential and potential energy

1 (a) What is the potential at a distance of r = 2.12 x 10 hydrogen atom?

-10

m from the nucleus of an

(b) What is the electric potential energy of an electron in electron volt at the said point? 1.

2.

When an electron moves from A to B along and electric field line in the figure below, the electric field does 3.94 x 10-19 J of work on it. What are the electric potential differences (a) VA - VB , (b) VC -VA, and (c) VC - VB ? A B

3.

C

Two infinite lines of charge are parallel to and in the same plane with the z – axis. One, of charge per unit length + λ , is a distance a to the right of this axis. The other, of charge per unit length - λ , is a distance a to the left of this axis. Sketch some of the equipotential surfaces due to this arrangement. Consider a point charge q = +1.0 µ C, point A at a distance d1 = 2.0m from q, and point B at a distance d2 = 1.0m (a) If these points are diametrically opposite each other, as in Fig:a, what is the electric potential difference VA – VB? (b) What is the electric potential difference if point A and B are located as in Fig: b

4.

B B

5.

6. 7.

8.

9.

d2 + d1 q Fig a

A

d2 q Fig b d1

A

Two large parallel conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electric force of 3.5 x 10-15 N acts on an electron placed anywhere in between the plates (uniform field). (a) Find the electric field at the position of the electron. (b) What is the potential difference between the plates? An infinite non-conducting sheet has a surface charge density σ = 10 µ C / m2 on one side. How far apart are two equi-potential surfaces whose potential differ by 50 V? A charge q is distributed uniformly throughout a spherical volume of radius R. (a) Setting V = 0 at infinity, Show that the potential at a distance r from the center , where r < R, is given by V = q(3R2 - r2)/8π∈oR3 A spherical drop of water carrying a charge of 30 pC has a potential of 500V at its surface (with V=0 at infinity). (a) What is the radius of the drop? (b) If two such drops of same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop? Two metal spheres are 3 cm in radius and carry charges of + 1.0 × 10 −8 C and -3.0 × 10 −8 C respectively, assumed to be uniformly distributed. If their centres are 2 m apart, calculate (a) the potential of the point halfway between their centres and (b) the potential on each sphere.

10.

For the following charge configuration show that V(r) for points on the vertical axis, 1  q 2qa   + 2  . Is this an expected result? assuming r > > a, is given by V = 4πε 0  r r  r -q

11.

a

+q

a

+q

P

In a certain situation, the electric potential varies along varies along the x axis as shown in the graph below. Determine the x component of electric field for the interval ab and bc.

V(x) in volt65bc4321da12345678x in m

12.

What is the net potential at point P due to four point charges , if V= 0 at infinity? +5q d -5q

-5q

P d +5q 13.

Following figure shows an electric dipole in a plane with its center at the origin of coordinate system. What is the work require to transfer a unit positive charge from point P to Q. Q +q

P -q

14.

Figure below shows a ring of outer radius R and inner radius r = 0.2R ; the ring has a uniform surface charge density σ. With V= 0 at infinity, find an expression fo the electric potential at a point p on the central axis of the ring, at a distance z = 2.0 R from the center of the ring. P

σ

R 15.

The potential at an axial point for a charged disk was shown to be σ V = a 2 + r 2 − r . From this result show that E for axial points is given by 2ε 0

(

E=

σ 2ε 0

 r 1 −  a2 + r 2 

)

   