The capacitance of the 150V battery is found to be 2.23 x 10⁻¹¹ F.
The capacitance of the battery is given by the relation,
C = ε₀A/d
Where,
ε₀ is electric permittivity,
A is the area of the electrode,
d is the distance between the electrode.
Here, in this case,
The diameter of the electrode is 4cm, so, area of both the electrode is,
A = 2 x π(2x10⁻⁴)
A = 1.26 x 10⁻³ m².
The distance between the electrode is 0.50 x 10⁻⁴ m.
Value of the electric permittivity is 8.85 x 10⁻¹² Fm⁻²
So, putting the values,
C = 8.85 x 10⁻¹² Fm⁻² x 1.26 x 10⁻³ m²/0.50 x 10⁻⁴ m.
C = 2.23 x 10⁻¹¹ F.
So, the capacitance of the battery is 2.23 x 10⁻¹¹ F.
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