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 Since a cube has 6 identical sides and the point charge is at the center problem1 - Charge in a Cube Q Q=3.76 nC is at the center of a cube. What is.

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Presentation on theme: " Since a cube has 6 identical sides and the point charge is at the center problem1 - Charge in a Cube Q Q=3.76 nC is at the center of a cube. What is."— Presentation transcript:

1  Since a cube has 6 identical sides and the point charge is at the center problem1 - Charge in a Cube Q Q=3.76 nC is at the center of a cube. What is the electric flux through one of the sides?

2 2 Norah Ali Al-moneef King Saud university Problem 2 : Two large thin metal plates with surface charge densities of opposite signs but equal magnitude of 44.27 ×10 –20 Cm –2 are placed parallel and close to each other. What is the field (i)To the left of the plates? (ii) To the right of the plates? (iii) Between the plates? Electric field exists only in the region between the plates. Therefore, (i), (ii) E = 0

3 A spherical shell of radius 10 cm has a charge 2×10 –6 C distributed uniformly over its surface. Find the electric field (a) Inside the shell (b) Just outside the shell (c) At a point 15 cm away from the centre problem3 : 3 Norah Ali Al-moneef King Saud university q = 2 ×10 –6 C,R = 0.1 m,r = 0.15 m (a) Inside the shell, electric field is zero

4 A 5 nC charge is placed inside of a cubic surface. The cube is measured to be 0.5 m per side. What is the flux through the cube? problem4 :

5 Ex) What is the electric flux through this disk of radius r = 0.10 m if the uniform electric field has a magnitude E = 2.0x10 3 N/C? 5 Norah Ali Al-moneef King Saud university problem5 :

6 Two infinite parallel sheets of charge Due to uniform surface charge density, surface of the sheet Region II Region I

7 Two infinite parallel sheets of charge Region III

8 The sheets have equal and opposite charge density IIIIII If, then In regions I and III, E = 0 In region II, Special Case

9 Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field (a)10.0 cm and (b)20.0 cm from the center of the charge distribution. problem6 :

10 A sphere of radius 8.0 cm carries a uniform volume charge density r = 500 nC/m 3. What is the electric field magnitude at r = 8.1 cm? problem7 : 10 Norah Ali Al-moneef King Saud university

11 A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of λ, and the cylinder has a net charge per unit length of 2 λ. From this information, use Gauss’s law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis. Problem 8:

12 A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 μ C. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder. Problem 9: 12 Norah Ali Al-moneef King Saud university

13 A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find (a)the charge density of each face of the plate and (b)the total charge on each face. Problem 10:

14 A point charge q = +2 μC is at the center of a sphere of radius 0.5 m. (a) Find the surface area of the sphere. (b) Find the magnitude of the electric field at points on the surface of the sphere. (c) What is the flux of the electric field due to the point charge through the surface of the sphere? (d) Would your answer to part (c) change if the point charge were moved so that it was inside the sphere but not at its center? (e) What is the net flux through a cube of side 1 m that encloses the sphere? (a) A = 4πr 2 A = 3.14 m 2 (b) E = 7.19×104 N/C (c) Φ= EA cos θ = EA φ = 2.26×10 5 N.m 2 /C (d) No cha nge in φ if q is inside sphere (e) Apply Gauss’s law; φ unchanged φ = 2.26×10 5 N.m 2 /C


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