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Lecture 6 Problems

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Example 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.

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**E = 0 since no charge inside the Gaussian surface**

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 (2) a < r < b we construct a spherical Gaussian surface of radius r Where equal zero. Why?

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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 Where r>c Region (3) b > r < c E=0 How?

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Example: 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 + , and the hollow cylinder has a net charge per unit length of +2 . 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.

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** linner = -l Also inner + louter = 2l thus louter = 3l**

(a) Use a cylindrical Gaussian surface S1 within the conducting cylinder where E=0 linner = -l Also inner + louter = 2l thus louter = 3l (b) For a Gaussian surface S2 outside the conducting cylinder

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Example: Consider a long cylindrical charge distribution of radius R with a uniform charge density r. 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 , and it encloses a charge By applying Gauss’s law we get, Thus radially outward from the cylinder axis 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.

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Example: 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 E = - E1 + (- E2) = - 2E

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**b) At point between the two sheets E = E1 + (- E2) = zero**

(c) At point to the right of the two parallel sheets E = E1 + E2 = 2E

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Example: Two large metal plates face each other and carry charges with surface density +s and -s respectively, on their inner surfaces as shown in figure 4.24. What is E at points (i) to the left of the sheets (ii) between them and (iii) to the right of the sheets?

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**E = 8x104N/C A = 0.25m2 q = 0.17x10-6C sigma = 0.68x10-6C/m2**

Example: 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 E = 8x104N/C A = 0.25m2 q = 0.17x10-6C sigma = 0.68x10-6C/m2

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**Example: An electric field of intensity 3. 5**

Example: 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

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Example: 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? Example: (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 m.

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Example: 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. Use to find (a) zero (b) 5.4x103N/C (c) 540N/C

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Example: 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? j = EA = E 4pr2 j = 6.3x103 N.m2/C j = q/eo q=5.5x10-8C

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**(1) A closed surface encloses a net charge of 2. 50 x 10-6 C**

(1) A closed surface encloses a net charge of 2.50 x 10-6 C. What is the net electric flux through the surface? In what direction is this net flux? Ans: x 105 N m2/C; outward (2) What is the charge per unit area, in coulombs per square meter, of an infinite sheet of charge if the electric field produced by the sheet of charge has a magnitude of 4.50 N/C ? Ans x C/m2 (3) The electric field in the region between a pair of oppositely charges plane parallel conducting plates, each 100 cm2 in area, is 7.20 x 103 N/C. What is the charge on each plate? Neglect edge effects. Ans x C

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**[6] A conducting sphere carrying a charge ‘q’ has a radius ‘a’**

[6] A conducting sphere carrying a charge ‘q’ has a radius ‘a’. It is inside a concentric hollow conducting sphere of inner radius ‘b’ and outer radius ‘c’. The hollow sphere has no net charge. Calculate the electric field for: a) r < a; b) a < r < b; c) b < r < c; d) r > c; e) What is the charge on the inner surface of the hollow sphere? f) What is the charge on the outer surface? [7] A small sphere whose mass is 0.60 g carries a charge of 3.0 x 10-9 C and is attached to one end of a silk fiber 8.00 cm long. The other end of the fiber is attached to a large vertical conducting plate, which has a surface charge of x C/m2 on each side. Find the angle the fiber makes with the vertical plate when the sphere is in equilibrium.

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From Chapter 23 – Coulomb’s Law

From Chapter 23 – Coulomb’s Law

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