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Lecture 6 Application (Gauss’s Law)

Lecture 6 Application (Gauss’s Law).

robert-burns
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Lecture 6 Application (Gauss’s Law)

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  1. Lecture 6 Application (Gauss’s Law) General Physics 2, Lec 6, By/ T.A. Eleyan

  2. [1] A solid conducting sphere of radius a has a net charge +2Q.  A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge –Q as shown in figure.  Using Gauss’s law find the electric field in the regions labeled 1, 2, 3, 4 and find the charge distribution on the spherical shell. Region (1) r < a To find the E inside the solid sphere of radius a we construct a Gaussian surface of radius r < a E = 0 since no charge inside the Gaussian surface Region (3) b > r < c E=0 How? General Physics 2, Lec 6, By/ T.A. Eleyan

  3. Region (2) a < r < b we construct a spherical Gaussian surface of radius r Region (4) r > c we construct a spherical Gaussian surface of radius r > c, the total net charge inside the Gaussian surface is q = 2Q + (-Q) = +Q Therefore Gauss’s law gives General Physics 2, Lec 6, By/ T.A. Eleyan

  4. [2] A long straight wire is surrounded by a hollow cylinder whose axis coincides with that wire as shown in figure.  The solid wire has a charge per unit length of +λ , and the hollow cylinder has a net charge per unit length of +2λ .  Use Gauss law to find (a) the charge per unit length on the inner and outer surfaces of the hollow cylinder and (b) the electric field outside the hollow cylinder, a distance r from the axis. (a) Use a cylindrical Gaussian surface S1 within the conducting cylinder where E=0 General Physics 2, Lec 6, By/ T.A. Eleyan

  5. (b) For a Gaussian surface S2 outside the conducting cylinder General Physics 2, Lec 6, By/ T.A. Eleyan

  6. [3] Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ.  Find the electric field at distance r from the axis where r < R. If we choose a cylindrical Gaussian surface of length L and radius r, Its volume is πr²L, and it encloses a charge ρπr²L. By applying Gauss’s law we get, radially outward from the cylinder axis Thus Notice that the electric field will increase as r increases, and also the electric field is proportional to r for r<R. For the region outside the cylinder (r>R), the electric field will decrease as r increases. General Physics 2, Lec 6, By/ T.A. Eleyan

  7. Two Parallel Conducting Plates • When we have the situation shown in the left two panels (a positively charged plate and another negatively charged plate with the same magnitude of charge), both in isolation, they each have equal amounts of charge (surface charge density s) on both faces. • But when we bring them close together, the charges on the far sides move to the near sides, so on that inner surface the charge density is now 2s. • A Gaussian surface shows that the net charge is zero (no flux through sides — dA perpendicular to E, or ends — E = 0). E = 0 outside, too, due to shielding, in just the same way we saw for the sphere. General Physics 2, Lec 6, By/ T.A. Eleyan

  8. Two Parallel Nonconducting Sheets • The situation is different if you bring two nonconducting sheets of charge close to each other. • In this case, the charges cannot move, so there is no shielding, but now we can use the principle of superposition. • In this case, the electric field on the left due to the positively charged sheet is canceled by the electric field on the left of the negatively charged sheet, so the field there is zero. • Likewise, the electric field on the right due to the negatively charged sheet is canceled by the electric field on the right of the positively charged sheet. • The result is much the same as before, with the electric field in between being twice what it was previously. General Physics 2, Lec 6, By/ T.A. Eleyan

  9. [4] Two large non-conducting sheets of +ve charge face each other as shown in figure.  What is E at points (i) to the left of the sheets (ii) between them and (iii) to the right of the sheets? We know previously that for each sheet, the magnitude of the field at any point is a) At point to the left of the two parallel sheets General Physics 2, Lec 6, By/ T.A. Eleyan

  10. b) At point between the two sheets (c) At point to the right of the two parallel sheets General Physics 2, Lec 6, By/ T.A. Eleyan

  11. [5] A square plate of copper of sides 50cm is placed in an extended electric field of 8*104N/C directed perpendicular to the plate.  Find (a) the charge density of each face of the plate General Physics 2, Lec 6, By/ T.A. Eleyan

  12. [6] An electric field of intensity 3.5*103N/C is applied the x axis.  Calculate the electric flux through a rectangular plane 0.35m wide and 0.70m long if (a) the plane is parallel to the yz plane, (b) the plane is parallel to the xy plane, and (c) the plane contains the y axis and its normal makes an angle of 40o with the x axis. (a) the plane is parallel to the yz plane (b) the plane is parallel to the xy plane The angel 90 c) the plane is parallel to the xy plane The angel 40 General Physics 2, Lec 6, By/ T.A. Eleyan

  13. [7] A long, straight metal rod has a radius of 5cm and a charge per unit length of 30nC/m.  Find the electric field at the following distances from the axis of the rod: (a) 3cm, (b) 10cm, (c) 100cm. General Physics 2, Lec 6, By/ T.A. Eleyan

  14. [8] The electric field everywhere on the surface of a conducting hollow sphere of radius 0.75m is measured to be equal to 8.90*102N/C and points radially toward the center of the sphere. What is the net charge within the surface? General Physics 2, Lec 6, By/ T.A. Eleyan

  15. [9] A point charge of +5mC is located at the center of a sphere with a radius of 12cm.  What is the electric flux through the surface of this sphere? [10] (a) Two charges of 8mC and -5mC are inside a cube of sides 0.45m.  What is the total electric flux through the cube? (b) Repeat (a) if the same two charges are inside a spherical shell of radius 0. 45 m. General Physics 2, Lec 6, By/ T.A. Eleyan

  16. (a) At 12 cm the charge in side the Gaussian surface is zero so the electric field E=0 [12] A solid copper sphere 15cm in radius has a total charge of 40nC.  Find the electric field at the following distances measured from the center of the sphere: (a) 12cm, (b) 17cm, (c) 75cm. (b) General Physics 2, Lec 6, By/ T.A. Eleyan

  17. [13] Two long, straight wires are separated by a distance d = 16 cm, as shown below. The top wire carries linear charge density 3 nC/m while the bottom wire carries -5 nC/m. 1- What is the electric field (including direction) due to the top wire at a point exactly half-way between the two wires? 2- Find the electric field due to the bottom wire at the same point, exactly half-way between the two wires (including direction). 3- Work out the total electric field at that point. General Physics 2, Lec 6, By/ T.A. Eleyan

  18. using Gauss’ Law and cylindrical Gaussian surface as Lec.5 page 15 General Physics 2, Lec 6, By/ T.A. Eleyan

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