SOLUTIONS MANUAL – INTRODUCTION TO
ELECTRODYNAMICS, 5TH EDITION BY
GRIFFITHS | ALL 12 CHAPTERS COVERED
,Contents
1 Vector Ana lysis 4
2 Electrosta tics 26
3 Pote ntia l 53
4 Ele ctric Fie lds In Matter 92
5 Ma gne tosta tics 110
6 Ma gne tic Fie lds In Matte r 133
7 Electrodyna m ics 145
8 Conserva tion La ws 168
9 Ele ctrom a gne tic Wa ves 185
10 Pote ntia ls And Fie lds 210
11 Radia tion 231
12 Ele ctrodyna m ics And Re la tivity 262
,Chapter 1
Vector Analysis
PROB LEM 1.1
(a) From The Diagram, |B + C| Cos Θ3 = |B| Cos Θ1 + |C| Cos Θ2. Multiply By |A|.
|A||B + C| Cos Θ3 = |A||B| Cos Θ1 + |A||C| Cos Θ2.
So: A·(B + C) = A·B + A·C. (Dot Product Is Distributiv e)
Similarly: B + C Sin Θ3 = B Sin Θ1 + C Sin Θ2. Mulitply By
A n̂. A B + C Sin Θ3 n̂ = A B Sin Θ1 n̂ + A C Sin Θ2 n̂ . | sin θ
If n̂ Is The Unit Vector Pointing Out Of The Page, It Follows That ` ˛¸ x` ˛¸ x
|B| c os θ1 |C| c os θ2
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross Product Is Distributiv e)
(b) For The General Case, See G. E. Hay’s Vector And Tensor Analysis, Chapter 1, Section 7 (Dot Product) And
Section 8 (Cross Product)
PROB LEM 1.2
The Triple Cross-Product Is Not In General Associative. For Example,
Suppos e A = B And C Is Perpendi cul ar To A, As In The Diagram.
= B
Then (B⇥C) Points Out-Of-The-Page, And A⇥(B⇥C) Points Down,
And Has Magnitude ABC. But (A⇥B) = 0, So (A⇥B)⇥C = 0 /=
A⇥(B⇥C).
PROB LEM 1.3
√ √
A = +1 x̂ + 1 Ŷ — 1 ẑ ; A =
3; B = 1 x̂ + 1 ŷ + 1 ẑ ; B = 3.
√ √
A·B = +1 + 1 — 1 = 1 = AB Cos Θ = 3 3 Cos Θ ⇒ Cos Θ = 3
1.
θ
Y
1 1 70
3
X
PROB LEM 1.4
The Cross-Product Of Any Two Vectors In The Plane Will Give A Vector Perpendicular To The Plane. For
Example, We Might Pick The Base (A) And The Left Side (B):
A = —1 x̂ + 2 ŷ + 0 ẑ ; B = —1 x̂ + 0 ŷ + 3 ẑ .
, x̂ Ŷ ẑ
A⇥B =
1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ .
1 03
This Has The Right Direction, But The Wrong Magnitude. To Make A Unit Vector Out Of It, Simply Divide By
Its Length:
√
|A⇥B| = 36 + 9 + 4 = 7. A⇥B = 7
n̂ = |A 7 .
⇥B| 7
PROB LEM 1.5
x̂ Ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(By cz — Bzcy ) (Bzcx — Bx cz ) ( Bx cy — By cx )
= x̂[Ay (Bx cy By cx) Az(Bzcx Bxcz)] + ŷ () + ẑ ()
(I’ll Just Check The X-Component; The Others Go The Same Way)
= x̂ (A y b x c y — Ay by cx — Azbzcx + Azbxcz) + ŷ ( ) + ẑ ().
B(A·C) — C(A·B) = [Bx (Ax cx + Ay cy + Az cz ) — Cx (Ax bx + Ay by + Az bz )] x̂ + () ŷ + () ẑ
= x̂ (A y b x cy + Azb x cz — Ay b y cx — Azb zcx ) + ŷ ( ) + ẑ ( ) . They Agree.
PROB LEM 1.6
A⇥(B⇥C)+B ⇥(C ⇥A)+C ⇥(A⇥B ) = B(A·C)— C(A· B)+C(A· B)— A(C· B)+A(B· C)— B(C· A) = 0.
So: A⇥(B ⇥C) — (A⇥B)⇥C = —B⇥(C⇥A) = A(B·C) — C(A·B ).
If This Is Zero, Then Either A Is Parallel To C (Including The Case In Which They Point In Opposite Directions,
Or One Is Zero), Or Else B·C = B·A = 0, In Which Case B Is Perpendicul ar To A And C (Including The Case B
= 0.)
Conclusion: A⇥(B⇥C) = (A⇥B)⇥C →⇒ Either A Is Parallel To C, Or B Is Perpendicular To A And C.
PROBLEM 1.7
= (4 x̂ + 6 ŷ + 8 ẑ ) — (2 x̂ + 8 ŷ + 7 ẑ ) = 2 —2
√
= 4 +4 + 1 = 3
2
3
— 23 3
PROB LEM 1.8
(a) A¯Y B̄ Y + A¯Z B̄ Z = (Cos Ay + Sin Az)(Cos By + Sin Bz) + ( Sin Ay + Cos Az)( Sin By + Cos Bz)
= Cos2 Ay b y + Sin Cos (Ay b z + Azb y ) + Sin2 Azb z + Sin2 Ay b y Sin Cos (Ay b z + Azb y ) +
2
Cos Azb z
= (Cos2 + Sin2 )Ay b y + (Sin2 + Cos2 )Azb z = Ay b y + Azb z. X
(b) (Ax) 2 + (Ay ) 2 + (Az) 2 = C3 Aiai = C3 3
Rijaj C3 Rik ak = Cj,K (Cirij Rik ) Aj Ak .
J=1
⇢
1 If J = K
This Equals A2 + A2 + A2 Provided C3 Rijrik =
Moreover, If R Is To Preserve Lengths For All Vectors A, Then This Condition Is Not Only Sufficient But
Also
Necessary.
2 2
For2Suppose A = (1, 0, 0). Then Cj,K (Ci Rijrik ) Ajak = Ci Ri1Ri1, And This Must Equal 1 (Since We
Want + + = 1). Likewis e, C3 R R = C3 R R = 1. To Check The Case J /= K, Choose A = (1, 1, 0).
Then We Want 2 = Cj,K ( Ci Rijrik ) Ajak = Ci Ri1 Ri1 + Ci Ri2 Ri2 + Ci Ri1Ri2 + Ci Ri2Ri1. But We Al ready
Know That The Fi rst Two Sums Are Both 1; The Third And Fourth Are Equa l, So Ci Ri1Ri2 = Ci Ri2Ri1 = 0,
ELECTRODYNAMICS, 5TH EDITION BY
GRIFFITHS | ALL 12 CHAPTERS COVERED
,Contents
1 Vector Ana lysis 4
2 Electrosta tics 26
3 Pote ntia l 53
4 Ele ctric Fie lds In Matter 92
5 Ma gne tosta tics 110
6 Ma gne tic Fie lds In Matte r 133
7 Electrodyna m ics 145
8 Conserva tion La ws 168
9 Ele ctrom a gne tic Wa ves 185
10 Pote ntia ls And Fie lds 210
11 Radia tion 231
12 Ele ctrodyna m ics And Re la tivity 262
,Chapter 1
Vector Analysis
PROB LEM 1.1
(a) From The Diagram, |B + C| Cos Θ3 = |B| Cos Θ1 + |C| Cos Θ2. Multiply By |A|.
|A||B + C| Cos Θ3 = |A||B| Cos Θ1 + |A||C| Cos Θ2.
So: A·(B + C) = A·B + A·C. (Dot Product Is Distributiv e)
Similarly: B + C Sin Θ3 = B Sin Θ1 + C Sin Θ2. Mulitply By
A n̂. A B + C Sin Θ3 n̂ = A B Sin Θ1 n̂ + A C Sin Θ2 n̂ . | sin θ
If n̂ Is The Unit Vector Pointing Out Of The Page, It Follows That ` ˛¸ x` ˛¸ x
|B| c os θ1 |C| c os θ2
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross Product Is Distributiv e)
(b) For The General Case, See G. E. Hay’s Vector And Tensor Analysis, Chapter 1, Section 7 (Dot Product) And
Section 8 (Cross Product)
PROB LEM 1.2
The Triple Cross-Product Is Not In General Associative. For Example,
Suppos e A = B And C Is Perpendi cul ar To A, As In The Diagram.
= B
Then (B⇥C) Points Out-Of-The-Page, And A⇥(B⇥C) Points Down,
And Has Magnitude ABC. But (A⇥B) = 0, So (A⇥B)⇥C = 0 /=
A⇥(B⇥C).
PROB LEM 1.3
√ √
A = +1 x̂ + 1 Ŷ — 1 ẑ ; A =
3; B = 1 x̂ + 1 ŷ + 1 ẑ ; B = 3.
√ √
A·B = +1 + 1 — 1 = 1 = AB Cos Θ = 3 3 Cos Θ ⇒ Cos Θ = 3
1.
θ
Y
1 1 70
3
X
PROB LEM 1.4
The Cross-Product Of Any Two Vectors In The Plane Will Give A Vector Perpendicular To The Plane. For
Example, We Might Pick The Base (A) And The Left Side (B):
A = —1 x̂ + 2 ŷ + 0 ẑ ; B = —1 x̂ + 0 ŷ + 3 ẑ .
, x̂ Ŷ ẑ
A⇥B =
1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ .
1 03
This Has The Right Direction, But The Wrong Magnitude. To Make A Unit Vector Out Of It, Simply Divide By
Its Length:
√
|A⇥B| = 36 + 9 + 4 = 7. A⇥B = 7
n̂ = |A 7 .
⇥B| 7
PROB LEM 1.5
x̂ Ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(By cz — Bzcy ) (Bzcx — Bx cz ) ( Bx cy — By cx )
= x̂[Ay (Bx cy By cx) Az(Bzcx Bxcz)] + ŷ () + ẑ ()
(I’ll Just Check The X-Component; The Others Go The Same Way)
= x̂ (A y b x c y — Ay by cx — Azbzcx + Azbxcz) + ŷ ( ) + ẑ ().
B(A·C) — C(A·B) = [Bx (Ax cx + Ay cy + Az cz ) — Cx (Ax bx + Ay by + Az bz )] x̂ + () ŷ + () ẑ
= x̂ (A y b x cy + Azb x cz — Ay b y cx — Azb zcx ) + ŷ ( ) + ẑ ( ) . They Agree.
PROB LEM 1.6
A⇥(B⇥C)+B ⇥(C ⇥A)+C ⇥(A⇥B ) = B(A·C)— C(A· B)+C(A· B)— A(C· B)+A(B· C)— B(C· A) = 0.
So: A⇥(B ⇥C) — (A⇥B)⇥C = —B⇥(C⇥A) = A(B·C) — C(A·B ).
If This Is Zero, Then Either A Is Parallel To C (Including The Case In Which They Point In Opposite Directions,
Or One Is Zero), Or Else B·C = B·A = 0, In Which Case B Is Perpendicul ar To A And C (Including The Case B
= 0.)
Conclusion: A⇥(B⇥C) = (A⇥B)⇥C →⇒ Either A Is Parallel To C, Or B Is Perpendicular To A And C.
PROBLEM 1.7
= (4 x̂ + 6 ŷ + 8 ẑ ) — (2 x̂ + 8 ŷ + 7 ẑ ) = 2 —2
√
= 4 +4 + 1 = 3
2
3
— 23 3
PROB LEM 1.8
(a) A¯Y B̄ Y + A¯Z B̄ Z = (Cos Ay + Sin Az)(Cos By + Sin Bz) + ( Sin Ay + Cos Az)( Sin By + Cos Bz)
= Cos2 Ay b y + Sin Cos (Ay b z + Azb y ) + Sin2 Azb z + Sin2 Ay b y Sin Cos (Ay b z + Azb y ) +
2
Cos Azb z
= (Cos2 + Sin2 )Ay b y + (Sin2 + Cos2 )Azb z = Ay b y + Azb z. X
(b) (Ax) 2 + (Ay ) 2 + (Az) 2 = C3 Aiai = C3 3
Rijaj C3 Rik ak = Cj,K (Cirij Rik ) Aj Ak .
J=1
⇢
1 If J = K
This Equals A2 + A2 + A2 Provided C3 Rijrik =
Moreover, If R Is To Preserve Lengths For All Vectors A, Then This Condition Is Not Only Sufficient But
Also
Necessary.
2 2
For2Suppose A = (1, 0, 0). Then Cj,K (Ci Rijrik ) Ajak = Ci Ri1Ri1, And This Must Equal 1 (Since We
Want + + = 1). Likewis e, C3 R R = C3 R R = 1. To Check The Case J /= K, Choose A = (1, 1, 0).
Then We Want 2 = Cj,K ( Ci Rijrik ) Ajak = Ci Ri1 Ri1 + Ci Ri2 Ri2 + Ci Ri1Ri2 + Ci Ri2Ri1. But We Al ready
Know That The Fi rst Two Sums Are Both 1; The Third And Fourth Are Equa l, So Ci Ri1Ri2 = Ci Ri2Ri1 = 0,
