general physics 2, lec 6, by/ t.a. eleyan 1 lecture 6 application (gauss’s law)

General Physics 2, Lec 6, By/ T.A. Eleyan

1

Lecture 6

Application(Gauss’s Law)


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[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 < aE = 0 since no charge inside the Gaussian surface

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


3

Region (2) a < r < bwe construct a spherical Gaussian surface of radius r

brawherer

QE

QrE

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Region (4) r > cwe 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

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QrE

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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

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0.0

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inner

encl

thus

Also

qdAE


5

(b) For a Gaussian surface S2 outside the conducting cylinder

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qdAE encl

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0

2

3

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)2(

.


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,

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.

radially outward from the cylinder axis


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 ) 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 2.

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.


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.


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]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 sheets02

E

0

21 2)(

E

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(c) At point to the right of the two parallel sheets

b) At point between the two sheets

0

21 2

E

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0)( 21 EEE


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[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

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/108

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Cq

qEA

mA

CNE


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[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.

CNmEA /5.8570cos 2

)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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]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.

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findtor

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/104.5)

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[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?

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q

echnetthefindtoNow

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23

2

105.5

arg

/.103.6

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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?

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/105.5 25

[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.

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/104.31085.8

)105108( 2512

66


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[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 .

(a) At 12 cm the charge in side the Gaussian surface is zero so the electric field E=0

outwardradiallyCNEr

qE

qrE

qEA

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encl

encl

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0

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2

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(b)


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[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.


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using Gauss’ Law and cylindrical Gaussian surface as Lec.5 page 15

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general physics 2, lec 6, by/ t.a. eleyan 1 lecture 6 application (gauss’s law)

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