Question

A parallel-plate capacitor has plates of area 0.12 m$$^2$$ and a separation of 1.2 cm. A battery charges the plates to a potential difference of 120 V and is then disconnected.A dielectric slab of thickness 4.0 mm and dielectric constant 4.8 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted?(b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab?

Answer 1

(a) Initially, the capacitance is :
$$C_0=\dfrac{ε_0A}{d}=\dfrac{(8.85\times10^{-12}C^2/N.m^2)(0.12 m^2)}{1.2\times10^{-2}m^2}=89pF$$
(b) Working through Sample Problem — “Dielectric partially filling the gap in a capacitor” algebraically, we find :
$$C=\dfrac{ε_0Aκ}{κ(d-b)+b}=\dfrac{(8.85\times10^{-12}C^2/N.m^2)(0.12m^2)(4.8)}{4.8)(1.2-0.40)(10^{-2}m)+(4.0\times10^{-3}m)}=1.2\times10^2pF.$$
(c) Before the insertion, $$q$$ = $$C_0$$$$V$$$$ (89 pF)(120 V)$$ = $$11 nC$$.
(d) Since the battery is disconnected, q will remain the same after the insertion of the slab,with $$q$$ = $$11 nC$$ .
(e) $$E=q/ε_0A=11\times10^{-9}C/(8.85\times10^{-12}C^2/N.m^2)(0.12m^2)=10kV/m$$
(f) $$ E'=E/κ=(10kV/m)/4.8=2.1kV/m.$$
(g) The potential difference across the plates is :
$$V$$ = $$E(d – b) + E'b$$ = $$(10 kV/m)(0.012 m – 0.0040 m)+ (2.1 kV/m)(0.40 × 10^{–3} m)$$ = $$88 V$$.

(h) The work done is :
$$W_{ext}=\Delta U=\dfrac{q^2}{2}\left( \dfrac{1}{C}-\dfrac{1}{C_0}\right) =\dfrac{(11\times10^{-9}C)^2}{2} \left( \dfrac{1}{89\times10^{-12}F}-\dfrac{1}{120\times10^{-12}F} \right) =-1.7\times10^{-7}J. $$