Responda:
1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5
Explicação:
Dado o seguinte:
Carga (q) = 3uC = 3 × 10 ^ -6C
Força elétrica (Fe) = 18N
Intensidade do campo elétrico (E) =?
1)
Lembre-se:
Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)
Fe = qE; E = Fe / q
E = 18N / (3 × 10 ^ -6C)
E = 6N / 10 ^ -6C
E = 6 × 10 ^ 6NC ^ -1
2)
Lembre-se:
E = kQ / r ^ 2
E = intensidade do campo elétrico
Q = carga de origem
r = distância de espera = 30cm = 30/100 = 0,3m
K = 9,0 × 10 ^ 9
6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2
9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09
Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9
Q = 0,06 × 10 ^ (6-9)
Q = 0,06 × 10 ^ -3
Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC
Answer:
a) P =392.4[Pa]; b) F = 706.32[N]
Explanation:
With the input data of the problem we can calculate the area of the tank base
L = length = 10[m]
W = width = 18[cm] = 0.18[m]
A = W * L = 0.18*10
A = 1.8[m^2]
a)
Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:
P = density * g *h
where:
density = 1000[kg/m^3]
g = gravity = 9.81[m/s^2]
h = heigth = 4[cm] = 0.04[m]
P = 1000*9.81*0.04
P = 392.4[Pa]
The force can be easily calculated knowing the relationship between pressure and force:
P = F/A
F = P*A
F = 392.4*1.8
F = 706.32[N]
Answer:
The charge resides on the outer surface = 
C
Explanation:
Surface area of cell 
Separation between two plate 
Dielectric constant 
Potential difference 
The capacitance of parallel plate capacitor in free space is given by,
Where 
permittivity of free space = 
The Capacitance of capacitor is increase by 
times when it placed in dielectric medium.
And we know that, 
So charge on the outer surface is given by,
