Ed eletricidade básica
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FIXA 17 EXER 1 PAG 3AB=d OP=x Q= 1*10^(-3)C d=0,002m x=4,0M q=5*10^(-4)C m=0,1kq
1/(4*PI()*ε0)= 9*10^9 Nm2/C2 AP=4+0,01m BP= 4-0,01m
Fap=1/(4*PI()*ε0)*Qa*q/AP2
Fap=9*10^9*(-1*10)^(-3)* 5*10^(-4) /(4+0,01)^2
Fap=-281,109N
Fbp=1/(4*PI()*ε0)*Qb*q/BP^2
Fbp=9*10^9*1*10^(-3)* 5*10^(-4) /(4-0,01)^2
Fbp= 281,39N
Fr= Fap+Fbp
Fr=-281,109+281,39
Fr= 0,281N
Fr= m*a a=0,281/0,1 a=2,81m/s^2
FIXA 17 EXER 2 PAG 4
AB=d OP=x Q= 1*10^(-3)C d=0,002m x=4,0M q=5*10^(-4)C m=0,1kq
1/(4*PI()*ε0)= 9*10^9 Nm2/C2 AP=4+0,01m BP= 4-0,01m Fap=-281,109N Fbp= 281,39N
Fr= Fap+Fbp
Fr=-281,109+281,39
Fr= 0,281N
Fr=E*q
E= Fr/q
E =0,281/5*10^(-4)
E =562 N/C
FIXA 18 EXER 1 PAG 5
Q =2*10^(-6) r=4m 1/(4*PI()*ε0)= 9*10^9 Nm2/C2
E=1/(4*PI()*ε0)*Q*x /(r^2+x^2)^(3/2)*i
E=[Q/(4*PI()*ε0)]*x*(r^2+x^2)^(-3/2) /u
Regra para determinar o máximo campo dE/dx=[Q/(4*PI()*ε0)]*[1*(r^2+x^2)^(-3 /2)+x*(-3/2) *(r^2+x^2)^(-3-1/2)*2x]=0
Q/(4*PI()*ε0)=0
(r^2+x^2)^(-3 /2)-3*x^2*(r^2+x^2)^(-5/2)=0
(r^2+x^2)^(-3 /2)=3*x^2*(r^2+x^2)^(-5/2)
Simplificando
1=3*x^2* (r^2+x^2)^(-1)
1=3*x^2/ (r^2+x^2)^ 1
3*x^2= r^2+x^2
3*x^2+x^2=r^2
x^2=r^2/2 x^2=4^2/2 x^2=16/2 x=8^(1/2) x=2,828m
FIXA 18 EXER 2 PAG 6
E=1/(4*PI()*ε0)*Q*x /(r^2+x^2)^(3/2)*i
Q =2*10^(-6) r=4m 1/(4*PI()*ε0)= 9*10^9 Nm2/C2
E =1/(4*PI()*ε0)*Q*x /x^3* i
E = 1/(4*PI()*ε0)*Q /x^2* i
FICHA 19 EXER 1 PAG 7
Q=5*10^(-6) C L=10 m A=4 m
1/(4*PI()*ε0)= 9*10^9 Nm2/C2 dE= 1/(4*PI()*ε0)* da/R^2 λ=dq/dx dq= λ*dx
R=L+A-x
dE=1/(4*PI()*ε0)*λ*dx/( L+A-x) ʃ dE= λ/(4*PI()*ε0)* ʃ(de 0 a L) dx/( L+A-x)^2
E= λ/(4*PI()*ε0)* ʃ(de 0 a 10) dx/( 10+4-x)^2
E= λ/(4*PI()*ε0)* ʃ(de 0 a 10) dx/( 14-x)^2
Substituição u=14-x du=0-dx dx=-du
E= λ/(4*PI()*ε0)* ʃ -du/u^2
E= -λ/(4*PI()*ε0)* ʃ u^(-2)* du
E= -λ/(4*PI()*ε0)* [ u^(-1)/- 1]
E= -λ/(4*PI()*ε0)* [-1/u ]
E= λ/(4*PI()*ε0)* 1/u
E= λ/(4*PI()*ε0)* [1/(14-x) ] (de 0 a 10) λ= Q/L
E= 1/(4*PI()*ε0)* Q/L *