A capacitor with a capacitance of 100 F is charged to a potential difference of 50 V . The charging battery is then disconnected and a dielectric of dielectric constant 2.5 is inserted. Determine the charge induced at a surface of the dielectric slab.

Step-by-step solution

Initial charge on the capacitor is given by Q = CV . Substituting the given values, Q = 100 F 50 V = 5000 C = 5 mC . When the battery is disconnected, the charge on the capacitor plates remains constant. The induced charge Q_i on the surface of the dielectric is given by the formula Q_i = Q ( 1 - 1 K ) . Substituting Q = 5 mC and K = 2.5 , we get Q_i = 5 ( 1 - 1 2.5 ) = 5 ( 1 - 2 5 ) = 5 3 5 = 3 mC .