A 15-?F air-filled capacitor is connected to a 50-V voltage source and becomes f

Question

A 15-?F air-filled capacitor is connected to a 50-V voltage source and becomes fully charged. The voltage source is then removed and a slab of dielectric that completely fills the space between the plates is inserted. The dielectric has a dielectric constant of 5.0. A) What is the capacitance of the capacitor after the slab has been inserted? B) What is the capacitance of the capacitor after the slab has been inserted?

Explanation / Answer

Without dielectric, in capacitor, field inside E = Q /(eo*A) Pot diff = V = E * d = Qd /(eo*A) C = Q/V = (eo*A) /d ----- (1) where d = plate separation, 4pi*eo = 1/9*10^9 With dielectric (k) these are E1 = Q / k (eo*A) Pot diff = V1 = E * d = Qd /k(eo*A) C1 = Q/V = k (eo*A) /d ----- (2) dividing C1 = k C = 5*15*10^-6 Farad = 75 micro Farad OR capacity increases by 5 times when filled fully (d thickness) with dielectric (k=5) what happens to Q and V (1) let us keep the Battery attached (voltage ON) when we insert dielectric plate >>> in this case both Q and C will vary in such a way to keep V = constant V = V1 = Q/C = Q1/C1 Q1 = (C1/C) * Q = k Q **** when battery is not removed but k inserted >> Q & C get k times V1 = 50 V, Q1 = 5 *15*10^-6*50 = 0.00375 Coulomb, C1=75 micro Farad YOUR CASE (2) >>> Battery removed when fully charged then V and C will vary in such a way that total charge remains same or Q = Q1 = CV = C1V1 V1 = (C/C1) V = (1/k) V here when battery removed and k inserted >> C will be k times (5C) but voltage will DROP by (1/5) its initial value V1 = 50 /5 = 10 Volts C1 = 5 *15 = 75 micro Farad

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