Problem Set 1 2s1112 Solutions Problem 1 – J. De Guzman Problem 2,3 – A. Tio Problem 4,5,6 – L. Paet
Problem 1: Vectors and Coordinate Systems o o o (a) (a) What What is d ⃗ N in along the green member if θ L =30 ? θ L =120 ? θ L =85 ? o θ L =30 : √ 2 , √ 2 , √ 3 > â L = < sin30co sin30cos45 s45 , sin30sin sin30sin45 45 , cos30 cos30 > =< 4 4 2 √ 2 , √ 2 , √ 3 >× < −1,0 √ 3 , √ 2 > ̂ X =< 1,0,0 ,0 > = < 0 , − d ⃗ N =â L×−a 4 4 2 2 4 o θ L =120 : √ 6 √ 6 1 â L = < sin120c sin120cos45 os45 , sin120si sin120sin45 n45 , cos120 cos120 >= < , ,− > 4 4 2 1 √ 6 √ 6 √ 6 1 ⃗ N = â L×−a ̂ X = < , , − > ×< −1,0,0 1,0,0 > = < 0 , , > d 4 4 2 2 4 o θ L =30 : â L = < sin85 0.704,0.704,0.08 ,0.087 7> sin85cos cos45 45 , sin85 sin85si sin45 n45 , cos85 cos85 >= < 0.704,0.704 ⃗ = â ×−a ̂ = <0.704,0.704,0.087> × <− 1,0,0> = < 0 ,−0.087 , 0.704 0.704 > d N
L
X
(b) Find Find the the angle angle θ L that would make the green member or d ⃗ N lie on the z + y =0 plane . θ L =? : We use the general form for â L = < cos cos 45sin 45sin θ L , sin sin 45sin 45sin θ L , cos θ L > . ⃗ = â × a⃗ = < cos cos 45sin 45sin θ L , sin sin 45sin 45sin θ L , cos θ L > × <− 1,0,0> d N L ̂ X ⃗ N = â L× a⃗ X = < 0,− cos θ L , √ 2 sin θ L > which should lie on the z + y =0 plane d ̂ 2 √ 2 sin θ → − y = z → −(−cos θ L )= L 2 −1 o → θ L = tan √ 2 =54.74 (c) Now, Now, find find the the θ L that would make the green member as close as possible to point (x,y,z) = (-1,2,3). ) =(−1,2,3 ) : θ L =? , ( w , y , z )=(− Since, d ⃗ does not have an x-component, we can easily deduce that the pseudo N
⃗˙ = <0,2,3> is the vector nearest to the point (-1,2,3). The pseudo d ⃗˙ is d N N parallel to the projection of a vector (-1,2,3) on the yz-plane. Now, similar to (b), we ⃗ N = â L× a⃗ X = < 0,− cos θ L , √ 2 sin θ L > . can use the general form of â L to get d ̂ 2 Remember, the most important component of the green member is its direction ⃗ which is directed by d ˙ = <0,2,3>. Thus, we can use the ratio of the y- and z N
components as basis for calculation.
→
y −(cos θ L) 2 = = 3 z 2 √ ( sin θ L ) 2
→ θ L = tan
−1
−
6
( 2 √ 2 )
=115.24o
(d) Transform the vector originating from the tip of your green member in (c) to point (x,y,z) = (-1,2,3) to spherical coordinates. ⃗˙ = <0,2,3> with magnitude | d ⃗ ˙ |= ( 22+ 3 2)= 13 . Again, we use psuedo d N
N
√
√
The green member is then represented by the following vector 3 6 9 ⃗ =( 3 ) < 0, 2 > =< 0, > d N √ 13, √ 13 √ 13, √ 13 6 → R⃗cart = < −1,2,3 >− < 0,
9
√ √
13, 13
> =< −1,0.336,0.504 >
originating from (0,2,3): −1 2 o o θ= tan ( )= 33.69 & ϕ= 90 3 ⃗ → R r =[(−1)( sin33.69cos90 )+ ( 0.336)( sin33.69sin90 )+ ( 0.504 )( cos33.69 )] â r ⃗r =[( 0.336 )( sin 33.69 )+ (0.504 )( cos33.69 )] â r = 0.606 â r R → R⃗θ=[(−1)( cos33.69cos90 )+ (0.336 )( cos33.69sin90 )+ (0.504 )(− sin33.69 )] â θ R⃗θ=[( 0.336 )( cos33.69 )+ ( 0.504)(− sin33.69 )] â θ =0 â θ ⃗ϕ=[(−1 )(− sin90 )+ ( 0.336)( cos90 )] â ϕ=[(−1)(− 1)+ 0 ] â ϕ= â ϕ → R → R⃗ ̂ θ+ â ϕ sph =0.606 â r + 0 a
Problem 2
F 1= forceonQ 1 ⃗
F 2= forceonQ 2 ⃗
F 3= forceonQ 3 ⃗
Then equilibrium requires that
F 1= F 2= F 3 ⃗
⃗
⃗
The two original charges are both negative which mean they would repel each other. The third charge has to be positive and has to lie somewhere between them in order to counteract their repulsion force. The forces acting on charges Q1, Q2, and Q3 are respectively:
̂ Q Q R ̂ Q Q R 21 1 2 31 1 3
F 1= ⃗
2
4 πϵ 0 R21
+
2
4 πϵ 0 R31
̂ Q Q R ̂ Q Q R 12 1 2 32 3 2
F 2= ⃗
2
4 πϵ0 R12
+
2
4 πϵ0 R32
=− x̂
324 e
2
+ x̂ 2
4 πϵ0 d 2
= x̂
324 e
− x̂ 2
4 πϵ0 d
9 e Q3 4 π ϵ0 x
2
36 e Q 3 2
4 π ϵ0 ( d − x )
̂ 13 Q 1 Q3 R ̂ 23 Q 2 Q 3 9 e Q3 36 e Q 3 R + =− + ̂ F 3= x x ̂ 2 2 2 2 4 πϵ 0 R13 4 πϵ 0 R23 4 πϵ 0 x 4 πϵ 0 ( d − x ) ⃗
Hence, equilibrium requires that
−324 e 2
d
+
9Q 3
x
2
= 3242 e − d
−9Q3 36Q3 = + ( d − x )2 x 2 ( d − x ) 2 36Q3
Solution of the above equation yields
Q3= 4 e , x =
d 3
Problem 3a
The strip of charge density ρS (C/m2) can be treated as a set of adjacent line charges each of charge ρ L = ρS dy and width dy. At point P , the fields of line charge at distance y and line charge at distance -y give contributions that cancel each other along the y-direction and add along the z-direction. For each such pair, 2 ρS dy cos θ d E = z ̂ 2 πϵ 0 R ⃗
with R = h/cosθ , we integrate from y = 0 to d/2 , which corresponds to θ = 0 to θ 0 =sin-1[ (d/2) / (h 2 + (d/2)2 )1/2 ]. Thus, d / 2
∫
E = ⃗
0
d E = z ̂ ⃗
ρS πϵ 0
d / 2
∫
0
ρS cos θ dy = z ̂ πϵ 2 0
For an infinitely wide sheet, θ 0 = π/2 and ρS E = z ̂ 2ϵ ⃗
0
θ0
∫
0
2 ρS cos θ h˙ = θ d z ̂ πϵ θ0 h cos 2 θ 0
Problem 3b
ρV dV
∫ 4π
E = ⃗
2
2
ϵ0
â r
2
r
2
2
2
r = x + y + [− z −( x + y )]
2
〈− x , − y ,− z −( x 2+ y 2 )〉 â r = 2 √ x + y 2+ [− z −( x 2 + y 2)] 2
∫ρ
V
∫∫
dV =
ρ f ( x , y )+ f y ( x , y )+ 1 dxdy R S √ x 2
2
=∫∫ R ρS √ 4 ( x 2+ y 2 )+ 1 dxdy
V dV ∫ 4 r 2 ar 0
E =
S 4 x 2 y 2 1 dxdy 〈− x , − y , − z − x 2 y 2〉 E =∫∫ R 2 2 2 2 2 3/ 2 [ x y [− z − x y ] ]
∫
E =
2 0
S 4 21 d d ∫0 [2− z − 22 ]3 / 2 − z − 2 a z 2
V / m
Problem 4
The electric field due to the plane charge is given by, −9 S 1× 10 E plane = a = a x 2 0 N 2 0
The net force force experienced by the plane charge due to the point charge has the same magnitude as the force experinced by the point charge due to the plane charge but of opposite direction. Hence, −9
1×10 F plane=−q E plane= 2 0
˙ −9 a =−2.82 ×10−7 a N 5×10 x x
Following the same argument for the line charge,
−2 ×10 E = a = a 2 2 4 −9
L
line
0
x
0
−2 ×10−9 ˙ −9 a =4.49 ×10−8 a N 5 ×10 F line =−q E line = x x 2 0 4
The net force eaperienced by the point charge is given then given by,
−7
−8
−7
F point =− F plane F line =−−2.82 ×10 4.49 ×10 =−3.27 ×10 a x N
Problem 5
2 = -5.312 µC/m s1 s2
= 5.312 µC/m2
|charge/mass| = |Q/m| = 10 µC/kg for both the coal & ash.
ASSUMPTIONS for coal/ash particles: Initial distance travelled: d x(t=0) = dy(t=0) = 0 Initial velocity v x(t=0) = vy(t=0) = 0 a.) E-field due to s1
s1 a x 2 0 −5.312x106 6 3.000x10 = ≈ − a a x x −12 2 8.854x10
E s = 1
N / C
E-field due to s2
s2 −a x E s = 2 0 2
6
=
5.312x10
−12
2 8.854x10
−a x ≈ −3.000x106 a x N / C
Total E-field
= E s1 E s2 E TOTAL ≈ − 6.000x106 a x
N / C
//
b.) Take a negatively charged particulate:
“-”
Fe
Fg
vertical motion only affected by gravity horizontal motion only affected by electric field For the vertical motion acceleration due to gravity: g = constant integrating once wrt. to time: vy(t) = g t + c 1 integrating twice: dy(t) = (g t2) / 2 + c1 t + c2 using initial conditions, vy(t=0) = g (0) + c 1
→
c1 = 0
dy(t=0) = (g (0)2) / 2 + c1 (0) + c2
→
c2 = 0
final equation for vertical motion: dy(t) = (g t2) / 2 for the particle, we want to know the time t1 it takes to fall 1m down.: dy(t1) = (g (t1)2) / 2 = -1m (-10 m/s2) (t1)2 = - 2m t1 = sqrt(0.2) secs.
For the horizontal motion
E = q E = m a E F q a E = E m −6
6
a E = −10x10 − 6x10 = 60 m / s integrating once wrt. to time t , v x t = a E t c 3 integrating twice , 2 a E t d x t = c 3 t c 4 2 using initial conditions , v x 0 = 0 = a E 0 c 3
2
a constant
c3= 0
2
d x 0 = 0 =
a E 0 2
c3 0 c 4
c 4=0
final equation for horizontal motion : 2 2 a E t 60 t d x t = m = 2 2
after time t1 , the negatively charged particle travelled a horizontal distance of :
0.2
2
d x t1 = 60
= 6m 2 also , after time t1 , the corresponding positively charged particle will have travelled 6min the opposite direction therefore , d TOTAL = 12 meters //
Problem 6
= 10x a x −5yx a y E 3
E y E x
= −
2
5yx
2
10x
3
=
d y d x
d y = y 2x d x d x d y 2 = − x y
−
integrating both sides , we get the general equation for the streamlines ln | x | =−2 l n | y | c
@ P 4,1,1 , ln 4
= − 2ln 1 c
c = ln 4
.⋅. equation for streamline passing through P is ln | x |
= −2 l n | y | l n | 4 | / /
@ Q 16,0.5,1 , ln 16
= − 2 ln 0.5 c
c = ln 4
.⋅. equation for streamline passing through Q is also ln | x |
= −2 l n | y | l n | 4 | / /