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

Answer:

K = 2 10⁻⁸ J

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 = $\frac{Q d_2}{ \epsilon_o A}$

we substitute the equation for Q

         ΔV₂ = $\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}$

         ΔV₂ = $\frac{d_2}{d_1} \ \Delta V_1$

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 $\frac{d_2}{d_1}$

we calculate

          K = 1 10⁻⁹  12  0.005/0.003

          K = 2 10⁻⁸ J