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Solutions1_2010

# Solutions1_2010 - Problem 1 A water droplet of radius 0.015...

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Problem 1 . A water droplet of radius 0.015 mm remains hanging stationary in the air. If the electric field of Earth is 100 N/C, how many excess electrons must the water droplet have? In which direction the electric field of Earth must be directed for that problem to make any sense? First of all, if the droplet hangs stationary in the air, then it means electrostatic force and gravitation force cancel each other. Also, since we are told that there are excess electrons (negatively charged) in the droplet, the only way we can have electrostatic force pointing up is to require electric field to be pointing DOWN. The number of excess electrons is N=q/e (where e= - 1.6x10 -19 C, charge of the electron) E q+m g= 0, Eq= - mg= Vg= -ρ4/3 π r 3 g N= -ρ4/3 π r 3 g /E e = -1000 kg/m 3 4/3 π (0.015 . 10 -3 m) 3 9.8 m/s 2 / 100N/C (-1.6x10 -19 C) =8.65x10 6 electrons. (Kg/m 3 )m 3 m/s 2 / (N/C) C = (kgm/s 2 )/N = dimensionless (N=kgm/s 2 as F=ma) Problem 2. An electric dipole is formed from two charges, +/-q, spaced 1 cm apart. The dipole is centered at the origin, oriented along the y-axis. The electric field strength at the point (x,y)=(0cm, 10cm) is 360 N/C. What is the charge q? What is the electric field strength at the point (x,y)=(10cm, 0cm)? mg F E =qE E Ne (0,10) (10,0) + - E+ E-

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First let’s consider the electric filed at the point (0,10). This field may be represented as a sum of two unequal components, one due to positive charge of the dipole, and another due to negative charge of the dipole. They partially cancel each other, as they point in different directions. ) 1 1 ( ) 10 , 0 ( 2 , 1 2 , 1 2 , 1 2 , 1 - + + - - + + - = + = r r q k r q k r q k E e e e where r 1,+ =9.5 cm=0.095 m and r 1,- =10.5 cm=0.105 m C m m C Nm C N r r k E q e 9 2 2 2 2 9 2 , 1 2 , 1 10 2 ) 105 . 0 ( 1 ) 095 . 0 ( 1 / 10 99 . 8 / 360 1 1 ) 10 , 0 ( - - + + = - = - = or 2 nC And now we can just use the textbook example (turned by 90 degrees) with a=0.5 cm and x=10 cm. (there was y in the example, but I will use x for the sake of consistency with my own figure). The electric field vector at the point (0,10) will be pointing down and C N m m C m C Nm x a aq k E e y / 8 . 179 ) ) 005 . 0 ( ) 1 . 0 (( 10 2 005 . 0 2 / 10 99 . 8 ) ( 2 2 / 3 2 2 9 2 2 9 2 / 3 2 2 = + = + = - + NB: It is also an acceptable solution to use the simplified formula for x>>a (y>>a in the textbook example) As long as the units in the above equations are properly crossed, the review is considered complete. Problem 3 : Two 10-cm charged disks face each other, 20 cm apart. The left disk is charged to -50 nC and the right disk is charged to 50 nC. What is the electric field (both magnitude and direction) at the midpoint between the disks? What would be the force (again, magnitude and direction) on a -1 nC test charge placed at the midpoint?
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