#5 please (1) A uniform electric field of magnitude 433 N/C pointing in the posi
ID: 2032652 • Letter: #
Question
#5 please
(1) A uniform electric field of magnitude 433 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.40 cm. (a) What is the work done by the field on the electron? (b) What is the change in potential energy associated with the electron? (c) What is the velocity of the electron? (2) A metal sphere of radius 5.00 cm is initially uncharged. How many electrons would have to be placed on the sphere to produce an electric field of magnitude 1.65 x 10' N/C at a point 7.56 cm from the center of the sphere? 3(a) When a 6.50-V battery is connected to the plates of a capacitor, it stores a charge of 28.0 HC. What is the value of the capacitance? (b) If the same capacitor is connected to a 17.5-V battery, what charge is stored? (4) Consider the following. (Let C,-40.00 ?F and C-34.00 ?F.) 6.00 ?F C, 9.00 V (a) Find the equivalent capacitance of the capacitors in the figure. (b) Find the charge on each capacitor. (c) Find the potential difference across each capacitor (5) Consider the following figure. 7.00 2 4.00 2 9.00 2 et (a) Find the equivalent resistance between points a and b in the figure. (R-| 70 ?) (b) Calculate the current in each resistor if a potential differene points a and b. e of 26.0 V is applied betweenExplanation / Answer
(1).(a) W = F • d
= EQ • d
= ( 433 N/C )( -1.602 × 10^-19 C )( - 0.0340 m )
= +2.35 × 10^-18 J
Work is positive because the force is in the same direction as motion.
(b) Since no other forces are acting on the electron, the work is the increase in kinetic energy and the decrease in potential energy:
-2.35 × 10^-18 J
(c) ½mv² = KE ( Kinetic Energy )
v² = 2KE / m
v² = 2( 2.35 × 10^-18 J ) / ( 9.11 × 10?31 kg )
v² = 5.15 × 10^12 m²/s²
v = 2.26 × 10^6 m/s
Since electrons have a negative charge, they are pulled in the opposite direction of the electric field. So the electron is moving to the left ( -x direction ).
(2). E = kQ / d^2 = 9 . 10^9 . Q / 0.076^2 = 1.65 . 10^5
Q = 1.03 . 10^–7 C
number of electrons = Q / e = 1.03 . 10^–7 / 1.6 . 10^–19 = 6.43 . 10^11
(3). a. C=Q/V
=28*10^-6÷6.5
C= 4.3×10^-6 F
b. Q= CV
= 4.3×10^-6×17.5
Q= 75.25×10^-6 C
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