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.

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C =

        C = ε₀

we solve for the charge (Q)

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q =

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          ΔV₂ =

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

we calculate

         K = 1 10⁻⁹  12    

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

where's the following? you gotta add it to know what the choices are

explanation: