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
