TEST 2 (PHY 250) Figure Figure P26.21

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Georgiana Morgan
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1 TEST 2 (PHY 250) 1. a) Write the efinition (in a full sentence) of electric potential. b) What is a capacitor? c) Relate the electric torque, exerte on a molecule in a uniform electric fiel, with the ipole moment of the molecule. ) What is the electrical strength of a ielectric? 2. ielectric slab of thickness an area is inserte into the space between the plates of a parallel-plate capacitor with spacing s an surface area (as in igure P26.59). Derive an expression for the capacitance of the system in terms of the given imensions. s igure Over a certain region of space, the electric potential epens on position accoring to the following function: 5x+3x 2 y+2yz 2. in the expression for the electric fiel vector in this region. 4. ro of length lies along the x axis with its left en at the origin an has a nonuniform charge ensity λ(x) αx, where α is a positive constant) a) What is the unit of α? b) alculate the electric potential of point. 5. our capacitors are connecte as shown in igure P a) in the equivalent capacitance between points a an b. b) Determine the voltage across the 20µ capacitor if a 15 potential ifference exits between these points (a an b). a 15 µ 3 µ 6 µ igure P µ b

2 - 1 - a) Electric potential (r) at position r is a number such that the electric potential energy U(r) of a particle with charge q place at this location woul be U(r) q (r) b) capacitor is an electrical element with two sies, calle plates, for which the potential ifference (voltage) between the plates is proportional to the charge Q transferre from one plate to the other Q (The proportionality constant is calle capacitance of the capacitor.) c) The electric torque t exerte on a molecule with the ipole moment p, in a uniform electric fiel E, is given by t p E (This equation efines the ipole moment.) ) The ielectric strength (of a ielectric) is efine as the maximum (magnitue of the) electric fiel that can exist in the ielectric without electric breakown.

3 - 2 - s +σ σ Solution 1 (more elegant) The charge ensity σ on the plates of a parallel plate capacitor is uniform. ccoring to the efinition of capacitance, the charge on the 1) Q σ σ plate is proportional to the potential ifference between the plates 2) Q ( ) or a uniform electric fiel (between the plates) it shoul not be ifficult to express that potential ifference in terms of the charge ensity. We can express the magnitue of the electric fiel vector in terms of the charge ensity an use the relationship between electric potential an electric fiel vector. or a parallel plate capacitor, the magnitue of the electric fiel vector is irectly proportional to the area charge ensity 3a) E σ ε 0 (in free space) (If we on t remember this expression it can be quickly erive from e electric fiel in the free space (between the plates) therefore has this value. Because the slab is a ielectric, the electric fiel insie the slab has a magnitue of 3b) E σ ε 0 (within the ielectric) Knowing the electric fiel vector at the consiere region allows us to fin the potential (ifference) 4) + E E s s E s + conucting slab free space E cos 180 E ( s )cos 180 E ( s + )

4 The rest is math. Beginning with equation (2) an consecutively eliminating unesire variables we fin Q σ σ ε ε E( s + ) σ( s + ) ( s ) Solution 2. We can consier one surface of the slab as the plates of two separate parallel plate capacitors connecte in series. rom the geometrical information we can express the capacitance of each capacitor in terms of the slab thickness 1) ε 1 0 ε 2 0 s Now we can fin the equivalent capacitance of the system 1 1 s + + ε ε ε s +σ σ +σ σ 0 + s ( )

5 - 3-5x + 3x y + 2yz 2 2 Both electric potential an the electric fiel vector escribe electric fiel. Therefore they are relate the electric fiel vector is opposite to the graient of the electric potential. pplying this relation to the given function yiels the answer. E,, x y z 2 2 [ 5 + 6xy,3x 2z, 4yz]

6 - 4 - y qλx x a) inear charge ensity relates charge with the linear imension of the object. Therefore the SI unit of charge ensity must be /m. oefficient α relates charge ensity with position. Therefore its unit must be /m 2. b) The contribution to the electric potential at point from a ifferential fragment of the ro epens on the charge of the piece an the istance to the point. rom the efinition of charge ensity it is possible to relate the charge of each piece with its length. Integration over the entire ro yiels the answer ro k k α ln 1 + ro q r αxx k x + 0 kα ( x ln( + x) ) 0

7 - 5 - a) The 15µ an the3 µ capacitors are connecte in series. Their equivalent capacitance is therefore & µ 15µ 3µ This system is connecte in parallel to the 6µ capacitance. Therefore the equivalent capacitance of the three capacitors &3& µ + 6µ 8.5µ The last (20µ) capacitor is in series with the rest yieling the equivalent capacitance of the entire system µ 1 20µ 6.0µ a 15 µ 3 µ Q 15 -Q 15 Q 3 -Q 3 Q 6 6 µ -Q 6 Q µ -Q 20 b b) Since the 20µ capacitor is connecte in series, the remaining capacitor is charge to the same value as the entire system. (Notice that the right-han sie of this capacitor is the right-han sie plate of the system.) Therefore Q 20 Q 6µ 15 90µ ollowing the efinition of capacitance, the voltage across this capacitor is Q 90µ 20µ

Where A is the plate area and d is the plate separation.

Where A is the plate area and d is the plate separation. DIELECTRICS Dielectrics an the parallel plate capacitor When a ielectric is place between the plates of a capacitor is larger for the same value of voltage. From the relation C = /V it can be seen that