Question

A parallel-plate capacitor is formed from two 1.6 cm -diameter electrodes spaced 2.8 mm apart. The electric field strength inside the capacitor is 6.0 × 10^6 N/C . What is the charge (in nC) on each electrode?

Express your answer using two significant figures.

Answer 1

The charge on each electrode of the parallel-plate capacitor is 534 nC.

Step-by-step explanation:

The charge on each electrode of a parallel-plate capacitor can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.

First, we need to find the capacitance of the capacitor using the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the electrodes, and d is the separation between the electrodes.

Substituting the given values (diameter = 1.6 cm, separation = 2.8 mm) into the formula, we have:

C = (8.85 × 10⁻¹² F/m)(π(0.08 m)²)/0.0028 m = 8.90 × 10⁻¹⁰ F

Finally, we can calculate the charge on each electrode using Q = CV:

Q = (8.90 × 10⁻¹⁰ F)(6.0 × 10⁶ N/C) = 5.34 × 10⁻⁴ C = 534 nC

Answer 2

12 nC

Step-by-step 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 .