Physics, 11.05.2021 15:40 nicko10

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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Answer from: emilypk1998

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;

$C_0 = \epsilon_0 \times \dfrac{A}{d}$

Where;

ε₀ = Constant

A = The area of the plates

With the new distance, 2·d, we get;

$C_{new} = \epsilon_0 \times \dfrac{A}{2\cdot d} = \dfrac{1}{2} \times \epsilon_0 \times \dfrac{A}{d} = \dfrac{1}{2} \times C_0$

Therefore;

$The \ new \ capacitance \ C_{new} = \dfrac{1}{2} \times C_0$

The potential difference, 'V', is given as follows;

$C = \dfrac{Q}{V}$

Therefore;

$V = \dfrac{Q}{C}$

Given that Q = Q₀, we get;

$V = \dfrac{Q_0}{\dfrac{1}{2} \times C_0} = 2 \times \dfrac{Q_0}{C_0} = 2 \times V_0$

∴ 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.

Answer from: Quest

stands for electromotive force and it is measured in volts. despite its name, it is not actually the force in the sense of being a "pushing" or "pulling" (not measured in newtons).an equation that relates emf, ℰ, to a battery's terminal voltage, v, current, i, and internal resistance, r, is ℰ = v + irin this case, the emf can be interpreted as a voltage—a difference in electric potential.i would not say that it would be considered an energy. if emf can be defined as the energy/work per a unit charge from a battery or other electric source, then it is not strictly energy.

Answer from: Quest

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: $k= \frac{1}{2}mv^2$ where m is the mass of the arrow and v its velocity. all the other options, instead, do not increase the kinetic energy of the system: - pulling farther back on the string : this will just increase the elastic potential energy of the system- holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)- decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)

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