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portielesc
11.05.2021 •
Physics
A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo, it has charge of magnitude Qo on its plates. It is then disconnected from the battery and the plates are pulled apart to a separation 2d without discharging them. After the plates are 2d apart, the new capacitance and the potential difference between the plates are: (Show all steps) [2 marks] a. ½ Co, ½ Vo b. ½ Co, 2Vo c. Co, Vo d. Co, 2Vo e. 2Co, 2Vo
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Ответ:
b. 1/2·C₀, 2·V₀
Explanation:
The capacitance on the parallel plate capacitor = C₀
The area of the plates = A
The voltage on the battery = V₀
The magnitude of the charge on the plate = Q₀
The new distance between the plates = 2·d
From an online physics source, we have;
Where;
ε₀ = Constant
A = The area of the plates
With the new distance, 2·d, we get;
Therefore;
The potential difference, 'V', is given as follows;
Therefore;
Given that Q = Q₀, we get;
∴ V = 2 × V₀
The new potential difference, V = 2·V₀
Therefore, after the plates are 2·d apart, the new capacitance and potential difference between the plates are;
1/2·C₀, 2·V₀.
Ответ:
Answer
It states that V/I = R where V is potential difference R is resistance and I is the current flowing across the terminals.Now when short circuit the resistance ...
Explanation: