A parallel-plate capacitor is formed from two 1.6 cm -diameter electrodes spaced 2.8 mm apart. The electric field strength insid e the capacitor is 6.0 × 10^6 N/C . What is the charge (in nC) on each electrode?
1 answer:
Answer:
12 nC
Explanation:
Capacity of the parallel plate capacitor
C = ε₀ A/d
ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate
Area of plate = π r²
= 3.14 x (0.8x 10⁻²)²
= 2 x 10⁻⁴
C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³
= 7.08 x 10⁻¹³
Potential difference between plate = field strength x distance between plate
= 6 x 10⁶ x 2.8 x 10⁻³
= 16.8 x 10³ V
Charge on plate = CV
=7.08 x 10⁻¹³ X 16.8 X 10³
11.9 X 10⁻⁹ C
12 nC .
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