2 JACOB LEWIS BOURJAILY ˜ x0 ) will automatically be a symmetric function if Therefore, we see that G(x, Z 1 F (x) − F (x0 ) = (G(x0 , y) − G(x, y)) da. S ∂Ω In particular, this will be true if we define F (x) by Z −1 F (x) = G(x, y)da. S ∂Ω We have shown that for any arbitrary Green’s function, G(x, x0 ), satisfyingRNeumann ˜ x0 ) ≡ G(x, x0 )− 1 boundary conditions there exists a Green’s function G(x, S ∂Ω G(x, y)da, which satisfies the same boundary conditions as G(x, x0 ) and has the property that ¶ Z µ ˜ y) ∂ G(x ˜ 0 , y) − G(x ˜ 0 , y) ∂ G(x, ˜ y) da = 0. G(x, ∂n ∂n ∂Ω ˜ x0 ) is From our derivation in part (a), this implies that the Green’s function G(x, symmetric. o´‘ π²ρ ²´’ δ²ι δ²ιξαι 2. Capacitance I a) We are to determine the capacitance of two large, flat, parallel conducting sheets of area A separated by a distance d. If we chose a Gaussian region that completely encloses one of the plates such that the edges are arbitrarily small, then the surface integral of the electric field will give E · A where E is the magnitude of the electric field. Notice that we have used the fact that the electric field will be non-vanishing only between the plates. Using Gauss’ law, was see that the surface integral is equal to the total charge contained within the region divided by ²0 . Specifically, we have that Q σ E·A= =⇒ E = , ²0 ²0 where σ is the charge density on the surface of one of the plates. The magnitude of the voltage difference between the two plates is equal to the line integral of the electric field from one plate to the other. Because we know that in the region between the two plates the electric field is independent of position, this will be simply ∆V = σd ²0 . Therefore, using the definition of capacitance, we see that A²0 ∴C= . d b) We are to determine the capacitance of two concentric conducting spheres with radii a and b where b > a. If we chose a Gaussian region that completely encloses the inner sphere, then the surface integral of the electric field will give E · A where E is the magnitude of the electric field and A = 4πr2 , the area of the boundary of the region. Using Gauss’ law, was see that the surface integral is equal to the total charge contained within the region divided by ²0 . Specifically, we have that Q Q =⇒ E = , E · 4πr2 = ²0 4π²0 r2 where a < r < b and Q is the charge on one of the spheres. The magnitude of the voltage difference between the two spheres is equal to the line integral of the electric field from one to the other. Specifically, Z b Z b Q 1 Q b−a ∆V = Ed` = dr = . 2 4π²0 a r 4π²0 ab a Therefore, using the definition of capacitance, we see that 4π²0 ab ∴C= . b−a PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 1 3 c) We are to determine the capacitance of two concentric conducting cylinders of length L with radii a and b where b > a. If we chose a Gaussian region that completely encloses one the inner cylinder, then the surface integral of the electric field will give E · A where E is the magnitude of the electric field and A = 2πrL, the area of the boundary of the region. Using Gauss’ law, was see that the surface integral is equal to the total charge contained within the region divided by ²0 . Specifically, we have that Q Q E · 2πrL = =⇒ E = , ²0 2π²0 Lr where a < r < b and Q is the charge on one of the cylinders. The magnitude of the voltage difference between the two cylinders is equal to the line integral of the electric field from one to the other. Specifically, µ ¶ Z b Z b Q dr Q b ∆V = Ed` = = log . 2π²0 L a r 4π²0 a a Therefore, using the definition of capacitance, we see that 2π²0 L ¡ ¢. ∴C= log ab 3. Capacitance II We are to approximate the capacitance per unit length of two parallel, cylindrical conductors with radii a1 and a2 which are separated by a distance d. Let us work within the coordinate system such that the center of the first cylinder, with radius a1 , is located at r = 0 and the second cylinder, with radius a2 , is located at r = d. Because the electric field is linear, we can consider the field caused by each of the conductors separately. Specifically, for points between the two cylinders, we can add the electric fields produced by each cylinder separately. We can determine the electric field per unit length induced by each cylinder by imagining a Gaussian region that completely encloses a unit length of either cylinder. Therefore, for a point collinear with the centers of each cylinder, we have that Q Q E= + . 2π²0 r 2π²0 (d − r) The magnitude of the voltage difference between the two cylinders is equal to the line integral of the electric field from one to the other. Specifically, ¶ Z d−a2 µ Q 1 1 ∆V = + dr, 2π²0 a1 r d−r · µ ¶ µ ¶¸ Q d − a2 a2 = log − log , 2π²0 a1 d − a1 µ ¶ (d − a2 )(d − a1 ) Q log , = 2π²0 a1 a2 µ 2 ¶ Q d ≈ log , 2π²0 a1 a2 µ ¶ Q d = log √ . π²0 a1 a2 Therefore, using the definition of capacitance per unit length, we see that π² ³ 0 ´. ∴C≈ log √ad1 a2