This preview shows page 1. Sign up to view the full content.
Unformatted text preview: maximum flux, Φ E = EA . Since A = π r 2 , we can write down the
magnitude of the field as
5 .2 x10 5 Nm 2 C
⎛ 0 .4 m ⎞
E = 4 .1 x10 6 N C
E= Question 2:
For A: With δ very small, all points on the hemisphere are nearly at distance R from the charge,
so the field everywhere on the curved surface is e 2 radially outwards. Then the flux is
this field strength time the area of the sphere: Solutions will be posted at www.prep101.com/solutions 11 ©Prep101 Phys1026 April Exam Solutions ( ) Q
⎛ k Q ⎞⎛ 1 ⎞
Φ CURVED = ⎜ e 2 ⎟ ⎜ ⎟ 4π R 2 =
Q (2π ) =
⎝ R ⎠⎝ 2 ⎠ For B:
The closed surface encloses zero charge, therefore Gauss’ Law gives:
Φ CURVED + Φ FLAT = 0
Φ FLAT = − Φ CURVED = − Q
2ε o Question 3:
If ρ is positive, the field must everywhere be radially outward. Choose as the Gaussian
surface a cylinder of length L and radius r, contained inside the charged rod. Its volume
is πr2L and it encloses charge ρ πr2L. The circular end caps have no electric flux through
them; there Φ E = ∫ E ⋅ dA = 0 . The curved surface has Φ E = ∫ E ⋅ dA = ∫ EdA cos θ , and E must be the same strength everywhere over the curved surface.
Then Φ E = ∫ E ⋅ dA = q εo ∫ becomes Φ E = E dA = ρπ r 2 L
εo Noting that 2πrL is the lateral surface area of the cylinder, E (2π r )L = E= ρπ r 2 L
2ε o radially away from the axis. Question 4:
Solution: For a large insulating sheet
2ε o = 2π k e σ So
Solutions will be posted at www.prep101.com/solutions 12 ©Prep101 ( Phys1026 April Exam Solutions )( ) E = 2π 8 .99 x10 9 Nm 2 C 2 8 .99 x10 −6 C m 2 = 5 .09 x10 5 N C E is perpendicular to the sheet.
The field just above the surface is given by E = σ/2ε, where σ = q/A. So we must find
which increases σ the most.
A. Increasing the charge by 4 μC increases the field by 1.4
B. This will have no effect – the spheres have the same charge.
C. Add +6 μC into the cavity, there must be -6 μC on the inner surface of the cavity to
prevent an electric field from appearing in the conductor. By conservation of charge,
there must now be 16 μC on the surface, increasing the field by 1.6
D. This will eliminate the electric field, by the same logic as C.
E. The area is πr2. Reducing the radius to 0.9 reduces the area to 0.81 of the old value,
increasing σ and the field by 1/0.81 = 1.23 C is the biggest increase.
A, C, D, B
A is the greatest, right next to a large charge. B is zero, inside a conductor. C is not zero,
as insulators allow some electric field. Whatever C is, it must be bigger than D which is
much further away, and D is still bigger than zero because all the charges present are the
Answer 4.7 a) This is outside the sphere, so by Gauss’ law we can treat it as a point charge located at
The total charge is
= 1 x 10-6 x π x (0.5)2
= 7.8 5 x 10-7 C
To cancel this out halfway between the two, th...
View Full Document
This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.
- Winter '14