Respuesta :
Answer:
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;
[tex]C_0 = \epsilon_0 \times \dfrac{A}{d}[/tex]
Where;
ε₀ = Constant
A = The area of the plates
With the new distance, 2·d, we get;
[tex]C_{new} = \epsilon_0 \times \dfrac{A}{2\cdot d} = \dfrac{1}{2} \times \epsilon_0 \times \dfrac{A}{d} = \dfrac{1}{2} \times C_0[/tex]
Therefore;
[tex]The \ new \ capacitance \ C_{new} = \dfrac{1}{2} \times C_0[/tex]
The potential difference, 'V', is given as follows;
[tex]C = \dfrac{Q}{V}[/tex]
Therefore;
[tex]V = \dfrac{Q}{C}[/tex]
Given that Q = Q₀, we get;
[tex]V = \dfrac{Q_0}{\dfrac{1}{2} \times C_0} = 2 \times \dfrac{Q_0}{C_0} = 2 \times V_0[/tex]
∴ 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.
At a distance of 2d, the potential difference will be twice the potential difference before while the capacitance will be half the value before.
Given to us
- The capacitance on the parallel plate capacitor = C₀
- The area of the plate = A
- The voltage on the battery = V₀
- The magnitude of the charge in the plate = Q₀
- The new distance between the plates = 2d
What is Capacitance?
We know capacitance can be written as,
[tex]C= \dfrac{\epsilon_0 A}{d}[/tex]
where C is the capacitance,
A is the area,
d is the distance and
ε₀ is the electrostatic constant,
Capacitance before,
[tex]C= \dfrac{\epsilon_0 A}{d}[/tex]
Capacitance Afterwards,
[tex]C= \dfrac{\epsilon_0 A}{2d}[/tex]
[tex]C=\dfrac{1}{2} \times \dfrac{\epsilon_0 A}{d}[/tex]
[tex]C=\dfrac{1}{2} \times C_0[/tex]
What is Voltage(Potential difference)?
We know that for the voltage we can write,
[tex]\rm Voltage = \dfrac{Charge}{Capacitance}[/tex]
Voltage before
[tex]V=\dfrac{Q_0}{C_0}[/tex]
Voltage Afterwards
[tex]V_{new} = \dfrac{Q_0}{C_{new}}[/tex]
[tex]V_{new} = \dfrac{Q_0}{\dfrac{1}{2} \times C_0}[/tex]
[tex]V_{new} = 2 \times\dfrac{Q_0}{C_0}[/tex]
[tex]V_{new} = 2 \times V_0[/tex]
Hence, At a distance of 2d, the potential difference will be twice the potential difference before while the capacitance will be half the value before.
Learn more about Capacitance:
https://brainly.com/question/12356566
