By Ilie, Carolina C

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Electromagnetism Solution Starting with the left-hand side ∇ × A⃗ = xˆ yˆ zˆ ∂ ∂x ∂ ∂y ∂ ∂z = [2z − ( −2xz )] yˆ + 4yzˆ = 2z(x + 1)yˆ + 4yzˆ . z 2 4xy −x 2z Now ( ) f ∇ × A ⃗ = 2xy 2 z 2(x + 1)yˆ + 4xy 3zzˆ . Here we have da ⃗ = dx′ dz nˆ where nˆ = xˆ + yˆ and n = from 2 so nˆ = 2 2 xˆ + 2 2 yˆ . Also we have dx′ = 2 dx with y = 1 − x . Now da ⃗ = ⎛ 2 2 ⎞ 2 dx dz ⎜ xˆ + yˆ ⎟ = dx dz xˆ + yˆ . 2 ⎠ ⎝ 2 ( ) Therefore, 1 1 ∫S f ( ∇ × A⃗ ) ⋅ da ⃗ = ∫ ∫ 0 0 1 1 ∫∫ = 0 ⎡ 2xy 2 z 2(x + 1)yˆ + 4xy 3zzˆ⎤ ⋅ (xˆ + yˆ )dx dz ⎣ ⎦ 2x(1 − x )2 z 2(x + 1)dx dz = 0 Next, we will solve the ∮ fA⃗ ⋅ dl⃗ term for the four segments.

0 ⎟ 0 0 ⎜ ⎟ ⎝ (i) ⎠ (ii) (iii) ∫ ∫ ∫ So, V (r ) = − k 2 x + y2 + z2 . 2 ( 2-14 ) Electromagnetism We can check this using JG E = − ∇V ⎧∂ ⎡ k ⎤ ⎤ ∂ ⎡ k 2 − ( x + y 2 + z 2 )⎥yˆ = − ⎨ ⎢ − ( x 2 + y 2 + z 2 )⎥xˆ + ⎢ ⎦ ⎦ ∂y ⎣ 2 ⎩ ∂x ⎣ 2 + ⎤ ⎫ ∂⎡ k 2 2 2 ⎢⎣ − ( x + y + z )⎥⎦zˆ⎬ ∂z ⎭ 2 2kx 2ky 2kz xˆ + yˆ + zˆ 2 2 2 JG = k (xxˆ + yyˆ + zzˆ ) = E . = JG c) E = k [2xzxˆ + z 2yˆ + (x 2 + 2yz )zˆ ] xˆ yˆ JG ∂ ∂ ∇×E = ∂x ∂y 2xz z 2 x 2 zˆ ∂ ∂z + 2yz ⎡∂ ⎡∂ ⎤ ∂ 2 ∂ 2⎤ = ⎢ ( x 2 + 2yz ) − z )⎥xˆ + ⎢ (2xz ) − x + 2yz )⎥yˆ ( ( ⎣ ∂z ⎦ ⎣ ∂y ⎦ ∂x ∂z ⎡∂ ⎤ ∂ + ⎢ ( z 2) − (2xz )⎥zˆ ⎣ ∂x ⎦ ∂y = k ⎡⎣ (2z − 2z )xˆ + (2x − 2x )yˆ + (0 − 0)zˆ⎤⎦ = 0.

Solution The force on q1 from q2 is given by JG 1 q1q2 1 q1q2 JG rˆ = r , F21 = 2 4πεo r 4πεo r 3 where JG r = 3xˆ + 4yˆ and r = 32 + 42 = 5. So JG F21 = 1 3q 2 q2 ⎛ 9 12 ⎞ ⎜ ˆ ˆ (3 4 ) + = x y xˆ + yˆ ⎟ . 3 4πεo 5 4πεo ⎝ 125 125 ⎠ The force on q1 from q3 is given by JG F31 = 1 q1q3 rˆ, 4πεo r 2 where r=4 and ˆ rˆ = y. 2-3 Electromagnetism So JG F31 = 1 ⎛ 2q 2 ⎞ q2 1 yˆ . ⎜ − 2 ⎟yˆ = − 4πεo ⎝ 4 ⎠ 4πεo 8 The force on q1 from q4 is given by JG F41 = 1 q1q4 rˆ, 4πεo r 2 where r=2 and ˆ rˆ = x. So JG F41 = 1 q2 q2 1 rˆ = xˆ .