Answer:
1.12μC
Explanation:
From the expression use to determine the capacitance of a capacitor, which is given below
[tex]C=q/v.........1\\C=E_{0}A/d-----2[/tex]
from the above since the distance,"d" is given is 1.1mm(0.0011m)
Area= 0.3m*0.08=0.024m
V=5.8KV
if we compare both equations above, we can do away with the capacitance
[tex]\frac{q}{v}=\frac{E_{0}A}{d} \\q=\frac{E_{0}Av}{d}\\[/tex]
if we substitute the value of the voltage, area, distance and the constant we arrive at
[tex]\\q=\frac{E_{0}Av}{d}\\ q=\frac{8.854*10^{-12}*5800*0.024}{0.0011}\\q=1.12*10^{-6}C\\[/tex]
Hence the magnitude of the charge is 1.12μC
1.12μC
(i) First let's get the capacitance (C) between the carpet and the shoe using the following relation;
C = A ε₀ / d ------------------------(i)
Where;
A = Area of the contact which is the area of the shoe = 30cm x 8cm = 240cm² = 0.024m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹² F/m
d = distance between the shoe and the carpet = 1.1mm = 0.0011m
Substituting for the values of A, d and ε₀ in equation (i) gives;
=> C = 0.024 x 8.85 x 10⁻¹² / 0.0011
=> C = 193.1 x 10⁻¹²F
(ii) Now calculate the magnitude of the charge
Remember that, Capacitance (C), magnitude of charge (Q) and potential difference (V) are related by the following;
Q = C x V ---------------------(ii)
This means that the quantity of charge transferred between two objects is the product of the potential difference (V) and the capacitance (C) between them.
Given;
potential difference, V = 5.8kV = 5.8 x 10³V
Found;
Capacitance, C = 193.1 x 10⁻¹²F
Substitute these values into equation (ii)
=> Q = 193.1 x 10⁻¹² x 5.8 x 10³
=> Q = 1.12 x 10⁻⁶C
=> Q = 1.12μC
Therefore the magnitude of charge that would have to be transferred between the carpet and the shoe is 1.12μC
