A capacitor is charged to a potential difference of 12v it delivers 40% of its stored energy to a lamp what is the final potential difference across the capacitor

Call the capacitance C.

Note the energy in a capacitor with voltage V is E = ½CV².

Initial energy = ½C (12) ² = 72C

40% of energy is delivered, so 60% remains. in the capacitor.

Remaining energy = (60/100) x 72C = 43.2C

If the final potential difference is X, the energy stored is ½CX²

½CX² = 43.2C

X² = 2 x 43.2 = 86.4

X = 9.3V