https://phet.colorado.edu/en/simulation/capacitor-lab-basicsThis is the link for the lab

D. Capacitor plate separation changed, while remaining connected to a Battery. Grab the Separation arrow (vertical green) and pull it up so that d = 10.0 mm. 1. Calculate the capacitance based on physical dimensions: C = E. A / d E. = 8.85E-12 The area is given in "mm?", convert this to A =im d = m. C= F (scientific) = PF. (3 significant figures for C) 2. Calculate the top plate charge, Q = C (scientific) = PC. The battery maintains 1.5 V on the plates. (3 significant figures for Q) 3. Calculate the field: use V = Ed, E = V/m. 4. Calculate the energy: PE = = J (scientific) = pJ. (3 significant figures for PE) E. Capacitor plate separation changed, while disconnected from the Battery. Grab the Separation arrow (vertical green) and move the separation back to d = 6.0 mm. This gets us back to case C. It puts a charge on the plates that equals the value in C2. Write this here, Q = _ C (sci notation). Move the switch to the off position (pivot to the right). The plates maintain their charge and voltage. Increase the separation to d = 10.0 mm. 1. Calculate the capacitance based on physical dimensions; C= E. A / d c. = 8.85E-12 The is the same as case D1. C = F (scientific) = PF. (3 significant figures for C) 2. But the voltage changes, while Q remains the same as it was at d = 6 mm. Calculate V from Q = CV, V = _V. 3. Calculate the field: use V = Ed, E= V/m. 4. Calculate the energy: PE = J (scientific) = pJ. Is this result similar to Example 17-12, p.487? Yes/No. (3 significant figures for PE)