d) Circle the equation(s) that correctly give the energy stored in a capacitor. 1Q? 2 0 Q (AV)² 1 C (AV) 10 1 2 Q?(AV) C (AV)² 2 0 1 Q 2 C2 e car) Q (AV) 1 c"(AV) e) Suppose you want the capacitor in Step 5 to have a capacitance of 13 x 10-12 F when the circular plates are separated a distance of 0.00050 m. What should the radius of the plates be?

y R The figure above represents a thin, circular ring of radius R= 5.8 m. The ring has a uniformly distributed charge Q=3.4 pC and point P is x = 14.8 m away from the center of the ring. (k = 9x10° V -m -c-1,p = 10-0) a) Calculate the electric potential value at point P and express your answer in SI units. Vp= V b) Calculate the electric field at point P. Ep=

Problem 5 Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation between the plates is d₂ = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V. The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa- ration of df = 8.00 mm. In the following, neglecting any fringing effects. (a) Will the new potential difference between the plates be larger, smaller, or the same compared to the initial potential difference of V₂ = 6.00 V? Explain. (Hint: Note that the charge will not change when the plates are pulled apart. Why is that?) (b) Find the potential difference Vf between the plates after the plates have been pulled to their new, larger separation df. (c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart. (d) To separate the plates, you will…