Question

There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.​

Answer 1

The kinetic energy acquired by a test charge of 1 nC moving through a 12 V potential difference is 12 nJ, as all the electric potential energy converts into kinetic energy.

Step-by-step explanation:

To determine the kinetic energy of a test charge transferred between the plates of an isolated capacitor after changing the plate separation, we must realize that the voltage across the capacitor plates remains constant when the capacitor is disconnected from the voltage source. The initial voltage (V) is 12 V, and capacitance (C) is not given but can be considered constant during the process as no charge is added or removed from the plates.

The electric potential energy (U) of a charge in an electric field is given by U = qV, where q is the charge and V is the potential difference. Since the charge is initially at rest, all the potential energy will convert to kinetic energy (KE) as the charge moves through the potential difference. So KE = qV.

Inserting the given values, we get KE = 1 nC × 12 V = (1 × 10^-9 C) × (12 J/C) = 1.2 × 10^-8 J or 12 nJ. Hence, the kinetic energy acquired by the test charge would be 12 nJ.

Answer 2

K = 2 10⁻⁸ J

Step-by-step explanation:

Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor

C = Q / ΔV

C = ε₀ A / d

ε₀ A / d = Q / ΔV

Q = ε₀ A ΔV / d (1)

indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,

as the power supply is disconnected and the capacitor is ideal the charge remains constant

in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1

ΔV₂ =
`(Q d_2)/( \epsilon_o A)`

we substitute the equation for Q

ΔV₂ =
`(d_2)/(\epsilon_o A) \ (\epsilon_o A \Delta V )/(d_1)`

ΔV₂ =
`(d_2)/(d_1) \ \Delta V_1`

in the third part we use the concepts of energy

starting point. Test charge near positive plate

Em₀ = U = q ΔV₂

final point. Test charge near negative plate

Em_f = K

energy is conserved

Em₀ = Em_f

q ΔV₂ = K

K = q ΔV₁
`(d_2)/(d_1)`

we calculate

K = 1 10⁻⁹ 12 0.005/0.003

K = 2 10⁻⁸ J