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Answer:</h2><h2>
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1.12μC
<h2>
Explanation:</h2><h2>
</h2>
<em>(i)</em> <em>First let's get the capacitance (C) between the carpet and the shoe using the following relation;</em>
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
<em>Substituting for the values of A, d and ε₀ in equation (i) gives;</em>
=> C = 0.024 x 8.85 x 10⁻¹² / 0.0011
=> C = 193.1 x 10⁻¹²F
<em>(ii) Now calculate the magnitude of the charge</em>
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.
<em>Given</em>;
potential difference, V = 5.8kV = 5.8 x 10³V
<em>Found;</em>
Capacitance, C = 193.1 x 10⁻¹²F
<em>Substitute these values into equation (ii)</em>
=> 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