Respuesta :
Answer:
The question is incomplete. It can be found in search engines. However, kindly find the complete question below.
Two 10.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes a. Right after the battery is disconnected? b. After insulating handles are used to pull the electrodes away from each other until they are 1.6 cm apart?
Recall the formula Q = CV
Where C = AE₀ / d
Where A = Area (πd²) and K = constant (k = 1)
Therefore, π(5²)E₀ / 0.5 = 50 πE₀ x 10⁻²
Therefore,
(a) After the battery is being removed, the charge on the capacitor will remain constant,
Therefore, charge on each electrode = 2.08 x 10 ⁻¹⁰ C opposite in sign
Hence, the electric field strength will be v / d = 3000 v/m
where potential difference = 15v
(b) After they are pulled apart such that distance (d) = 1.6 cm
the charge still remains constant
Therefore, the charge on each capacitor plate (electrode) will be the same
Hence, Q₁ = Q₂ = C₁V₁ = C₂V₂
Therefore,
A₁E₀ X 15 / d₁ = A₁E₀ V₂ / d₂ : where V₂ = 30 V
Therefore,
Electric Field = V₂ / e
= 30 / 1 x 10⁻²
= 3000v/m
