Evaluación 2 Buendia gonzalez

CAPACITANCIA

PROBLEMA 1:
VA= 110V
Q1 y Q2= 1200 µC C=Q/V= 1200x10 -6/ 110= 10.9 µf

PROBLEMA 2:
C= ((1)(0.34)(8.85x10⁻¹²))/((2x10⁻³) )= 1.5 nF
Q= CV= (1.5x10-9) (200)= 300 nC
E= V/d= 200/ 2x10-3= 100 x10 3

PROBLEMA 3:
V= Q/C= 800x10 -6 / 40x 10 -6= 20 V

PROBLEMA 4:
A= Cd/KEo= (5x10-9) (0.3x103-3)/ (1) (8.85x10-12)= 0.169 mm2= 169.49 m2
Q= cv= (5X10-9) (400)= 2X10-6 = 2µc

PROBLEMA 5:
C= K A Eo / d= (1) (0.6x10-2) (8.85x10-12) / (4x10-3)= 13.27 pF
C= (5) (0.6x10-2) (8.85x10-12) / (4x10-3) = 66.37 pF
E= V/d= (300) / (4x10-3) = 75x10 -3

PROBLEMA 6:
C= KAEo/ d= (7.5) (0.3x-2) (8.55x10-12)/ (4x10-3) = 49.78 pF
Q= CV= (49.78x10-2) (800) = 39.82 nC
E= V/d= (800) / (4x10-3) = 200x10-3

PROBLEMA 7:
C= KAEo / d= (2) (0.8x10-2) (8.85x10-12) / (4x10-3) = 35.4 pF
C= (1) (0.8x10-2) (8.85x10-12) / (4x10-3) = 1.7 Pf

PROBLEMA 8:
A= Cd/ K Eo= (2x10-9) (15x10-6) / (5) (8.85x10-12) = 6.77x10-4
D= V/ E= (3x103) / (200x106) = 15x10-6 m

PROBLEMA 9:
C= KAEo/ d= (4.3) (4x10-2) (8.85x10-12) / (2x10-3) = 7.61x10-6 F
Q= CV= (7.61x10-6) (100) = 7.6x10-4C

PROBLEMA 10
C= = KAEo/ d= (1) (5x10-2) (8.85x10-12) / (3x10-3) = 1.47x10-10 F
Q= CV= (= 1.47x10-10 F) (240) = 35.4 nC
E= V / d= (240) / (3x10-3)= 80x10-3

POTENCIAL ELECTRICO

PROBLEMA 1:
VA= K Q1/ r + K Q2/ r= (9x109) (8.40x10-6) / (6x10-2) + (9x109) (-2x10-9) / (6x10-2)
VA= 1.25x106V

PROBLEMA 2:
VA= (9x109) (-12x10-6) / (80x10-3) + (9x109) (3x10-6) / (80x10-3)
Va= -1.01x10-6 V

PROBLEMA 3:
VA= (9x109) (45x10-9) / (28x10-3) + (9x109) (-9x10-9) / (40x10-3)
VA= 12.43x103 V

PROBLEMA 4:
A= (9x109) (90x10-6) / (26x10-3)= 31.15x106 V
B= (9x109) (90x10-6) / (68x10-3) = 11.91x106 V
V= 19.24x106 V

PROBLEMA 5:
W= F d F=qd= (6x10⁻⁶) (8x10⁻²)=4.8x10⁻⁷
F=qd= (-2x10⁻⁶) (8x10⁻²)=-1.67x10⁻⁷
W=4.8x10⁻⁷-1.67x10⁻⁷= 3.13x10⁻⁷J

PROBLEMA 6:
D= V/E= (1200) / (4x105) = 3x10-3 m
VA= K Q/ r
Q= VA K/ r= (1200) / (3x10-3) = 400x103 C

PROBLEMA 7:
Vc=(9x〖10〗^9 )(8x〖10〗^(-6) )/(10x〖10〗^(-2) )+(9x〖10〗^9 )(-8x〖10〗^(-6) )/(10x〖10〗^(-2) )=0V
VD=(9x〖10〗^9 )(-8x〖10〗^(-6) )/(2x〖10〗^(-2) )=-3.6x〖10〗^6 V
F=qd=(-8x10-6)(8x10-6)=6.4x10-7J

PROBLEMA 8:
F=qE= (4x10⁻⁶)(6x10⁴)=240x10⁻³N
W=qEd= (30x10⁻³)(6x10⁴)(4x10⁻⁶)=7.2x10⁻³J
E) V=(9x〖10〗^9 )(4x〖10〗^(-6) )/(30x〖10〗^(-3) )=1.2x〖10〗^6 V

PROBLEMA 9:
Vc=(9x〖10〗^9 )(6x〖10〗^(-12) )/(4x〖10〗^(-2) )+(9x〖10〗^9 )(-6x〖10〗^(-12) )/(16x〖10〗^(-2) )=1.01V
Vb=(9x〖10〗^9 )(6x〖10〗^(-12) )/(10x〖10〗^(-2) )+(9x〖10〗^9 )(-6x〖10〗^(-12) )/(22x〖10〗^(-2) )=294.54x〖10〗^(-3) V
Va=(9x〖10〗^9 )(6〖x10〗^(-12) )/(8x〖10〗^(-2) )+(9x〖10〗^9 )(-6x〖10〗^(-12) )/(4x〖10〗^(-2) )=-675x〖10〗^(-3) V

PROBLEMA 10:
Va=(9x〖10〗^9 )(12x〖10〗^(-6) )/(4x〖10〗^(-2) )+(9x〖10〗^9 )(-12x〖10〗^(-6) )/(14x〖10〗^(-2) )=1.92x〖10〗^6 V
Vb=(9x〖10〗^9 )(12x〖10〗^(-6) )/(6x〖10〗^(-2) )+(9x〖10〗^9 )(-12x〖10〗^(-6) )/(4x〖10〗^(-2) )=-900x〖10〗^3 V
Vc=(9x〖10〗^9 )(12x〖10〗^(-6) )/(10x〖10〗^(-2) )+(9x〖10〗^9 )(-12x〖10〗^(-6) )/(10x〖10〗^(-2) )=0V

ENERGIA POTENCIAL

PROBLEMA 1:
Ep=KQq/r ((9x10⁹) (6x10⁻⁹)(80x10⁻⁶))/((50x10⁻³))=86.46x10⁻³J

PROBLEMA 2:
EpA=KQq/r (9x10⁹)(-9x10⁻⁶)/((40x10⁻³))=-2.02x10⁻³J
EpB=(9x10⁹)(-9x10⁻⁶)/((60x10⁻³))=-1.35x10⁻⁶J
Ep= ((9x10⁹)(-9x10⁻⁶)(-3x10⁻⁹))/((60x10⁻³))= -4.05x10⁻³J
Dif. Ep. EpA+EpB-Ep= (-2.02x10⁻³)+ (-1.35x10⁻⁶)-(-4.05x10⁻³)=2.02x10⁻³J

PROBLEMA 3:
Ep=KQq/r= r=KQq/Ep= (9x10⁹) (-7x10⁻³) (-3x10⁻⁹)/ ((60x10⁻³))=3.15mm
F=qE= (3x10⁻⁹) (2.72x10⁶)=8.16X10⁻³N/C

PROBLEMA 4:
Ep1=KQq/r=(9x10⁹)(6x10⁻⁶)(16x10⁻⁶)/((30x10⁻³) )=28.8J
Ep2=KQq/r=(9x10⁹)(6x10⁻⁶)(16x10⁻⁶)/((5x10⁻³) )=172.8J
Ep1-Ep2=172.8J-28.8J =144J

PROBLEMA 5:
Ep1=KQq/r=(9x10⁹)(6x10⁻⁶)(3x10⁻⁹)/((8x10⁻³) )=2.025x10⁻³J
Ep2=KQq/r=(9x10⁹)(6x10⁻⁶)(3x10⁻⁹)/((20x10⁻³) )= 8.1x10⁻³J
Dif. Pot.= Ep1-Ep2=2.025x10⁻³+ 8.1x10⁻³=6.075x10⁻³J

PROBLEMA 6:
Ep=KQq/r=rEp=KQq=r=KQq/Ep
r= (9x10⁹)(-7x10⁻¹²)(-12x10⁻¹²)/((9x10⁻¹²) )=84x10⁻³M

PROBLEMA 7:
Ep=KQ1q/r= Q1Q2=r Ep/K=√((Ep r)/K )=√(((4.5x10⁻³)(38x10⁻³))/(9x10⁹)) =134.74Nc

PROBLEMA 9:
Ep=KQq/r=rEp=Q1Q2kr=KQ1Q2/Ep= (9x10⁹)(5x10⁻⁹)(3x10⁹)/((75x10⁻²) )=18x10⁻²J
F=d/Ep=(9x10⁹)(18x10⁻²)/((75x10⁻²) )=2.16nN