A capacitor with a capacitance of 100 F is charged to a potential difference of 50 V . Subsequently, the charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Determine the new potential difference between the plates.

Step-by-step solution

Since the battery is disconnected after charging, the charge on the capacitor remains constant. Initial charge on the capacitor is Q = CV When a dielectric of dielectric constant K is inserted, the new capacitance becomes C' = KC . The new potential difference V' across the capacitor is given by: V' = Q C' = CV KC = V K Substituting the given values V = 50 V and K = 2.5 : V' = 50 2.5 = 20 V