Electric Fields Notes

9646 H2 PHYSICS

Lecture Notes

Chapter 12

ELECTRIC FIELDS Content • • • • •

Concept of an electric field Force between point charges Electric field of a point charge Uniform electric fields Electric potential

Learning Outcomes Candidates should be able to: (a) show an understanding of the concept of an electric field as an example of a field of force and define electric field strength as force per unit positive charge. (b) represent an electric field by means of field lines. (c) recall and use Coulomb's law in the form F = Q1Q2/4πεor2 for the force between two point charges in free space or air. (d) recall and use E = Q/4πεor2 for the field strength of a point charge in free space or air. (e) calculate the field strength of the uniform field between charged parallel plates in terms of potential difference and separation. (f) calculate the forces on charges in uniform electric fields. (g) describe the effect of a uniform electric field on the motion of charged particles. (h) define potential at a point in terms of the work done in bringing unit positive charge from infinity to the point. (i)

state that the field strength of the field at a point is numerically equal to the potential gradient at that point.

(j)

use the equation V = Q/4πεor for the potential in the field of a point charge.

(k) recognise the analogy between certain qualitative and quantitative aspects of electric and gravitational fields.

Nanyang Junior College

1

9646 H2 PHYSICS

Lecture Notes

Coulomb’s Law

F

+

Q1

Q2

+

F

r

Coulomb’s law states that the magnitude of the electrostatic force F that a point charge Q1 exerts on another point charge Q2, is proportional to the product of their charges and inversely proportional to the square of their separation r2. Q1 Q2 r2 Q Q F =k 12 2 r F∝

F=

1 Q1 Q2 4πε o r 2

where εo = 8.85 x 10-12 C2 N-1 m-2 or F m-1 is the permittivity of free space. Note: • F is an attractive force if Q1 and Q2 are unlike charges, and Q2 – – Q1 a repulsive force if Q1 and Q2 are like charges. F F • The permittivity of free space εo is usually taken to be equal Q1 to that of air. – Q2 + • The S.I. unit for charge is the Coulomb (C). F F • This law is applicable for point charges. For a uniformly charged conductor sphere, the law is applicable only if the separation of the centres of the spheres is large when compared to their radii.

qc

Example 1

+

qA (+7.0 µC), qB (−7.0 µC) and qC (+9.0 µC) are 3 charges located at the corners of an equilateral triangle of side 20 cm. Calculate the magnitude and direction of the net force on qC due to the other charges.

qA

qB

+

Nanyang Junior College

–

2

9646 H2 PHYSICS

Lecture Notes

Electric Field

An electric field is the region where an electrostatic force is felt by a charged body placed in the region. The field could be set up by a single charged body or by a collection of charged bodies.

F1

q1

+Q

+

q3

+

F3

+ q2 F2 • •

Small test charges q1, q2 and q3 experiences electrostatic forces F1, F2 and F3 respectively when placed in the electric field set up by charge Q. If the charge of Q is very large, there would be a “strong” electric field set up around it. Any test charge placed near Q would then experience a large electrostatic force.

Electric Field Strength

Supposed a very small positive test charge q0 is placed at a particular point in a electric field and it experiences an electrostatic force F. We define the electric field strength E at that point as follows: The electric field strength at a point in an electric field is defined as the electrostatic force per unit positive charge acting on a test charge placed at that point.

E=

F q0

Note: • The unit of E is N C-1 or V m-1. • E is a vector since F is a vector. • If q is positive, E and F would be in the same direction ⇒ positive charge moves in direction of E if released. • If q is negative, E and F would be in the opposite directions ⇒ negative charge moves opposite to direction of E if released. Nanyang Junior College

3

9646 H2 PHYSICS

Lecture Notes

Electric Field Lines

To visualize the electric field around a charged body, we can draw electric field lines to represent the direction of force on a small positive test charge placed in the field.

+

Properties of electric field lines: • • • • •

They begin from positive charges and end on negative charges, or at infinity. No two field lines can cross each other since the direction of E at a point is unique. The number of lines entering or leaving a charge is proportional to the magnitude of the charge. The number of field lines per unit cross sectional area is proportional to E. The stronger the field, the closer the lines are. In a uniform field, the lines are parallel and equally spaced. The field lines are perpendicular to the surface of the charge.

+

+

2 equal charges, one positive and one negative

+

2 equal positive charges

+ + + + + + +

+

-

+ 2 plane parallel plates

Nanyang Junior College

2 unequal charges, one positive and one negative

4

9646 H2 PHYSICS

•

Lecture Notes

For a non-linear electric field line, the electric field E is the tangent to the electric field line at each point. EQ EP

Q P electric field line

Electric Field Strength due to a point charge

Q

q0

+

E

r Consider a positive point charge Q in free space. If we placed a positive test charge qo, at a distance r away from Q, then by Coulomb’s Law, the electrostatic force that Q exerts on qo is 1 Q qo . F= 4πε o r 2 Thus the electric field strength of the field set up by Q is given by E = E=

F . Therefore, qo

1

Q 4πε o r 2

The direction of E is the same as that of the force on a positive charge: away from Q if Q is positive, and toward Q if Q is negative.

+

Q

q0 r

E

–

Q

E

q0

r

Note: • E decreases with distance from the point charge according to the inverse square law. • Field due to isolated charge is non-uniform but has spherical symmetry. It has the same values when viewed at any direction at equal distance from the point charge. • If charge Q is positive, then the field direction will radiate outwards, and if Q is negative, then it points inwards to the charge.

Nanyang Junior College

5

9646 H2 PHYSICS

Lecture Notes

Example 2

Q1

Consider 2 point charges located at the corners of a right-angled triangle where Q1 = +5.0 µC, Q2 = -2.0 µC and r = 0.10 m. Find the resultant electric field strength at point X.

X

r

+

r

_

Q2

Electric Field Strength between parallel charged plates

For two parallel plates having opposite charge, there is a uniform electric field between the plates. The electric field strength E at any point within the plates is given by d

E=

-

+ + + + + + + + + + +

V d

where V is the potential difference between the plates (V) and d, the separation of the plates (m).

+

V

–

Charge Accelerated Parallel to the Uniform Field E

An electron ejected at the negatively charged plate A with negligible kinetic energy (u = 0) will be accelerated to a final speed v before striking positively charged plate B. d

A -

u=0 _

FE = eE

– V

Nanyang Junior College

B + + + + + + + + + + +

+

6

9646 H2 PHYSICS

Lecture Notes

Since the electric field strength between parallel charged plates is uniform, the electrostatic force acting on the electron, and hence its acceleration will be constant. Applying Newton’s 2nd Law of Motion, ΣF = me a FE = me a a=

FE eE eV V = (since FE = eE ) = (since E = ) me me me d d

Applying Kinematics concepts: v 2 = u 2 + 2as  eV  v2 = 0 + 2 d  me d  2eV me

v=

Example 3 Two charged, parallel, flat conducting surfaces are spaced d = 1.00 cm apart and produce a potential difference V = 625 V between them. An electron is projected from one surface directly toward the second. Determine the initial speed of the electron if it comes to rest just at the second surface.

Charge Projected Perpendicularly into a Uniform Field E An electron is projected with a horizontal velocity u to enter a uniform electric field E perpendicularly as shown below: ++++++++++++++++++++++++++++

FE _

u

y

_

x

FE = eE = me ay ay =

eE eV = me me d

- - - - - - - - - - - - - - - - - - - -

The electron path inside the uniform electric field is a parabolic one. In general, if an electron is projected at an angle to a uniform electric field, the ensuing motion will be along a parabolic path. Nanyang Junior College

7

9646 H2 PHYSICS

Lecture Notes

Example 4 An electron is projected horizontally with a speed of 2.0 × 107 m s-1 into the uniform electric field set up between the two parallel charged plates spaced 35 cm apart as shown. It is deflected through a vertical distance of 10 cm in 1.0 × 10-7 s before emerging from the electric field. ++++++++++++++++++++++++++++

_

u

v

_

- - - - - - - - - - - - - - - - - - - -

Calculate (i) the vertical component of the velocity v, and (ii) the angle of deflection of the electron from the horizontal direction just before it emerges from the field. (iii) The electric field strength E, and (iv) The potential difference V between the two plates

Nanyang Junior College

8

9646 H2 PHYSICS

Lecture Notes

Electric Potential The electric potential V at a point in an electric field is defined as the work done per unit positive charge to bring a test charge from infinity to the point by an external agent. V is the electric potential energy per unit positive charge at that point in the field.

V=

U q

Note: • The unit of V is J C−1 or V. • V is a scalar. Thus, potential at a point due to two or more charges can be added or subtracted directly. • Any free positive charge placed in an E field will move from a point of higher to lower potential. • V decreases in the direction of E. The potential at infinite distance from an electric field is usually taken to be zero, but it is sometimes convenient to choose the earth to be the reference zero potential. • U = qV ⇒ ∆ U = q ∆V • Absolute values of V and U are meaningful only if we have defined a reference zero.

Potential due to a point charge It can be shown that for the field set up by a point charge Q, the electric potential at a point A located r from Q is given by

V=

1 Q 4πε o r

Q

A

r Note: • If Q is positive, then the electric potential at point A is positive. • If Q is negative, then the electric potential at point A is negative.

If several point charges Q1, Q2, etc. are at distances r1, r2, etc. from a point A, then the net potential at A can be obtained by scalar addition of V1, V2, etc. That is,

A

Q1

r3

r1

Q3

r2

VA = V1 + V2 + V3 + …….. Q2

Nanyang Junior College

9

9646 H2 PHYSICS

Lecture Notes

Example 5 Find the potential at points A and B, due to two small charges Q1 and Q2, 1.0 m apart in air and carrying charges +2.0 × 10-8 C and –2.0 × 10-8 C respectively. A

1.0 m

B

Q1

1.0 m

1.0 m

1.0 m

+

_

Q2

Electric Potential Energy m

Consider a ball falling freely under the influence of gravity. At different heights, the ball has different gravitational potential energies. As it falls, it loses potential energy and gains kinetic energy. We say that the ball has potential energy due to the gravitational field of the earth.

h

PE lost = mgh

A charged particle that is in an electric field experiences an electric force. If it is free to move, it will accelerate and gain K.E. We say that the system comprising the electric field and the charged particle together has electric potential energy U. The amount of potential energy the system has depends on the position of the charged particle in the field and the strength E of the field. Suppose q is a very small positive charge and Q is also positive. When released from rest at point A, q moves towards point B (which has a lower potential compared to A). In doing so, q loses electric potential energy and gains kinetic energy. Loss in electric potential energy U

Q

A

B

q

= UA – UB = qVA – qVB = q(VA – VB)

Thus, gain in KE = q ∆V

Nanyang Junior College

10

9646 H2 PHYSICS

Lecture Notes

Conversely, if q is moved from B to A at constant velocity by an external agent, there would be a gain in electric potential energy equal to q ∆V. Refer back to Section on Charge Accelerated Parallel to the Uniform Field E, Applying Energy concepts: In moving from A to B, Gain in Kinetic Energy = Loss in Electric Potential Energy 1 1 mev 2 − meu 2 = eV 2 2 1 mev 2 = eV Since u = 0 2 2eV v= me

Example 6 Solve Example 3 using energy considerations.

Example 7 An alpha particle (which consists of 2 protons and 2 neutrons) moves directly toward a gold nucleus, which has 79 protons and 118 neutrons. The alpha particle slows and then comes to a momentary stop, at a centre−to−centre separation r of 9.23 x 10−15 m, before it begins to move back along its original path. Because the gold nucleus is much more massive than the alpha particle, we can assume the gold nucleus does not move. Determine the kinetic energy K of the alpha particle when it was initially far away.

Nanyang Junior College

11

9646 H2 PHYSICS

Lecture Notes

Equipotential Surfaces Equipotential surfaces are surfaces on which the potential is the same at all points. The diagrams below are two examples of equipotential surfaces (dotted lines) and E field lines (bold lines):

+

E field due to point charge

Uniform E field due parallel plates

Note: • No work is done in moving a charge between any two points on an equipotential surface. • Equipotential surfaces are always at right angles to electric field lines. • The surface of a charged conductor of any shape is an equipotential surface. • The spacing between the equipotential surfaces will be closer where the field is stronger.

Electric Potential Gradient The negative rate of change of electric potential V in the direction of increasing r, or the potential dV gradient – , at a point in the field is equal to the component of the field strength Er in the dr direction of r at that point. dV Er = − dr Note: dV • is a vector and has units V m −1 or N C−1. dr • The negative sign shows that the potential decreases along the direction of the E field.

Example 8 2.0 cm

2.0 cm 2.0 cm

What is the value of the electric force exerted on a charge of +5.0 µC at point P in the field as shown in the diagram? P +

100 V

Nanyang Junior College

40 V

-20 V

-80 V

12

9646 H2 PHYSICS

Lecture Notes

Comparison between Gravitational and Electric Fields Gravitational Field

Electric Field

1

Origin of forces

Due to mass interaction

Due to charge interaction

2

Nature of force

Conservative and attractive.

Conservative and can be attractive or repulsive.

3

Quantitative Law

Newton’s Law of Universal Gravitation: mm FG = G 1 2 2 r

Coulomb’s Law: 1 Qq FE = 4πε 0 r 2

4

Field Strength

Force per unit mass: Fg (in N kg-1) g= m

Force per unit positive charge: F E = E (in N C-1) q

5

Field set up by isolated mass/charge

6

Potential

7

Potential due to isolated mass/charge

Φ=−

8

Potential due to multiple mass/charge

Φ net = −

g=

GM r2

E=

Work done per unit mass in bringing it from infinity to the particular point in a gravitational field.

GM (in J kg-1) r

Work done per unit positive charge in bringing it from infinity to the particular point in an electric field. V=

10

Potential energy of a 2 mass - 2 charge system Relation between field and potential

Nanyang Junior College

UG = −

g=−

1 Q (in J C-1) 4πε 0 r

1 Q1 1 Q2 GM1 GM 2 GM3 + − − −
Vnet = 4πε 0 r1 4πε 0 r2 r1 r2 r3 +

9

Q 4πε 0 r 2

GMm r

dΦ dr

1 Q3 +
4πε 0 r3

UE =

1 Qq 4πε 0 r

E=−

dV dr

13