xy 2 e 2z dx dy dz = 8 3 (1 e 4 ) = 2.62 mc. 12 x2 y 3 e 2z 2 m 2 m 2 m Figure P4.1: Cube of Problem 4.1.

Transcription

1 Problem 4.1 A cube m on a side is located in the first octant in a Cartesian coordinate system, with one of its corners at the origin. Find the total charge contained in the cube if the charge density is given by ρ v = xy e z (mc/m 3 ). Solution: For the cube shown in Fig. P4.1, application of Eq. (4.5) gives Q = ρ v dv = V x= ( ) 1 = 1 x y 3 e z y= z= x= y= z= xy e z dx dy dz = 8 3 (1 e 4 ) =.6 mc. z m m y m x Figure P4.1: Cube of Problem 4.1.

2 Problem 4.5 (a) ρ s = ρ s cosφ (C/m ) (b) ρ s = ρ s sin φ (C/m ) (c) ρ s = ρ s e r (C/m ) Find the total charge on a circular disk defined by r a and z = if: (d) ρ s = ρ s e r sin φ (C/m ) where ρ s is a constant. Solution: (a) (b) (c) (d) Q = a π r ρ s ds = ρ s cosφ r dr dφ = ρ s r= φ= a π Q = ρ s sin r φ r dr dφ = ρ s r= φ= = ρ sa 4 a π a sinφ π =. ( ) 1 cosφ dφ ) π ( φ sinφ a π a Q = ρ s e r r dr dφ = πρ s re r dr r= φ= [ = πρ s re r e r] a = πρ s [1 e a (1+a)]. a π Q = ρ s e r sin φ r dr dφ r= φ= a π = ρ s re r dr sin φ dφ r= φ= = ρ s [1 e a (1+a)] π = πρ s [1 e a (1+a)]. = πa ρ s.

3 Problem 4.6 If J = ŷ4xz (A/m ), find the current I flowing through a square with corners at (,, ), (,, ), (,, ), and (,, ). Solution: Using Eq. (4.1), the net current flowing through the square shown in Fig. P4.6 is I = J ds = (ŷ4xz) (ŷ dx dz) = ( x z ) = 16 A. S x= z= y= x= z= z m J y x m Figure P4.6: Square surface.

4 Problem 4.1 A line of charge of uniform density ρ l occupies a semicircle of radius b as shown in Fig. P4.1. Use the material presented in Example 4-4 to determine the electric field at the origin. z x b ρ l y Figure P4.1: Problem 4.1. Solution: Since we have only half of a circle, we need to integrate the expression for de 1 given in Example 4-4 over φ from to π. Before we do that, however, we need to set h = (the problem asks for E at the origin). Hence, de 1 = ρ lb 4πε E 1 = = ˆrρ l 4πε b dφ π φ= ( ˆrb+ẑh) (b + h ) dφ 3/ h= de 1 = ˆrρ l 4ε b.

5 Problem 4.11 A square with sides of m has a charge of 4 µc at each of its four corners. Determine the electric field at a point 5 m above the center of the square. Solution: The distance R between any of the charges and point P is R = = 7. E = Q [ R1 4πε R 3 + R R 3 + R 3 R 3 + R ] 4 R 3 = Q [ ˆx ŷ+ẑ5 ˆx ŷ+ẑ5 4πε (7) 3/ + (7) 3/ + ˆx+ŷ+ẑ5 (7) 3/ + ˆx+ŷ+ẑ5 ] (7) 3/ 5Q 5 4 µc = ẑ (7) 3/ = ẑ πε (7) 3/ = (V/m) = ẑ51. (kv/m). πε πε

6 z P(,,5) Q 3 (-1,-1,) R 3 R 4 R 1 R Q (-1,1,) Q 4 (1,-1,) Q 1 (1,1,) y x Figure P4.11: Square with charges at the corners.

7 Problem 4. Given the electric flux density determine (a) ρ v by applying Eq. (4.6). D = ˆx(x+y)+ŷ(3x y) (C/m ) (b) The total charge Q enclosed in a cube m on a side, located in the first octant with three of its sides coincident with the x-, y-, and z-axes and one of its corners at the origin. (c) The total charge Q in the cube, obtained by applying Eq. (4.9). Solution: (a) By applying Eq. (4.6) ρ v = D = x (x+y)+ (3x y) =. y (b) Integrate the charge density over the volume as in Eq. (4.7): Q = D dv = dx dy dz =. V x= (c) Apply Gauss law to calculate the total charge from Eq. (4.9) Q = D ds = F front + F back + F right + F left + F top + F bottom, F front = (ˆx(x+y)+ŷ(3x y)) (ˆx dz dy) y= z= x= = (x+y) dz dy = z (y+ 1 ) y= z= y = 4, x= z= y= F back = (ˆx(x+y)+ŷ(3x y)) ( ˆx dz dy) y= z= x= = (x+y) dz dy = zy = 8, y= z= x= z= y= F right = (ˆx(x+y)+ŷ(3x y)) (ŷ dz dx) x= z= y= ( ) 3 = (3x y) dz dx = z x= z= x 4x = 4, y= y= z= z= x=

8 F left = (ˆx(x+y)+ŷ(3x y)) ( ŷ dz dx) x= z= y= = (3x y) dz dx = x= z= y= z ( ) 3 x F top = (ˆx(x+y)+ŷ(3x y)) (ẑ dy dx) x= z= z= = dy dx =, x= z= z= F bottom = (ˆx(x+y)+ŷ(3x y)) (ẑ dy dx) x= z= z= = dy dx =. x= z= z= Thus Q = D ds = =. z= x= = 1,

9 Problem 4.5 The electric flux density inside a dielectric sphere of radius a centered at the origin is given by D = ˆRρ R (C/m ) where ρ is a constant. Find the total charge inside the sphere. Solution: Q = D ds = S π π θ= φ= ˆRρ R ˆRR sinθ dθ dφ R=a π = πρ a 3 sinθ dθ = πρ a 3 cosθ π = 4πρ a 3 (C).

10 Problem 4.8 If the charge density increases linearly with distance from the origin such that ρ v = at the origin and ρ v = 4 C/m 3 at R = m, find the corresponding variation of D. Solution: Hence, b =. ρ v (R) = a+br, ρ v () = a =, ρ v () = b = 4. ρ v (R) = R (C/m 3 ). Applying Gauss s law to a spherical surface of radius R, D ds = ρ v dv, S D R 4πR = V R R 4πR dr = 8π R4 4, D R = 5R (C/m ), D = ˆRDR = ˆR5R (C/m ).

11 Problem 4.3 A square in the x y plane in free space has a point charge of +Q at corner (a/,a/), the same at corner (a/, a/), and a point charge of Q at each of the other two corners. (a) Find the electric potential at any point P along the x-axis. (b) Evaluate V at x = a/. Solution: R 1 = R and R 3 = R 4. y -Q a/ Q R 3 R 1 -a/ a/ P(x,) x R 4 R -Q -a/ Q V = Q 4πε R 1 + Figure P4.3: Potential due to four point charges. Q 4πε R + Q 4πε R 3 + Q = Q ( 1 1 ) 4πε R 4 πε R 1 R 3 with ( R 1 = x ) a ( a ), + ( R 3 = x+ ) a ( a ). + At x = a/, R 1 = a,

12 R 3 = a 5, V = Q ( πε a ) =.55Q 5a πε a.

13 Problem 4.33 Show that the electric potential difference V 1 between two points in air at radial distances r 1 and r from an infinite line of charge with density ρ l along the z-axis is V 1 = (ρ l /πε )ln(r /r 1 ). Solution: From Eq. (4.33), the electric field due to an infinite line of charge is Hence, the potential difference is E = ˆrE r = ˆr ρ l πε r. r1 r1 ˆrρ l V 1 = E dl = r r πε r ˆr dr = ρ l ln πε ( r r 1 ).

14 Problem 4.35 For the electric dipole shown in Fig. 4-13, d = 1 cm and E = 4 (mv/m) at R = 1 m and θ =. Find E at R = m and θ = 9. Solution: For R = 1 m and θ =, E = 4 mv/m, we can solve for q using Eq. (4.56): E = qd 4πε R 3(ˆRcosθ + ˆθsinθ). Hence, ( ) qd E = = 4 mv/m at θ =, 4πε q = 1 3 8πε d = 1 3 8πε 1 =.8πε (C). Again using Eq. (4.56) to find E at R = m and θ = 9, we have E =.8πε 1 4πε 3 ( ˆR()+ ˆθ) = ˆθ 1 4 (mv/m).