A parallel-plate capacitor has plates of area 0.19 m2 and a separation of 1.50 c
ID: 2237318 • Letter: A
Question
A parallel-plate capacitor has plates of area 0.19 m2 and a separation of 1.50 cm. A battery charges the plates to a potential difference of 140 V and is then disconnected. A dielectric slab of thickness 4 mm and dielectric constant 4.8 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? ________________ pF (b) What is the capacitance with the slab in place? _______________ pF (c) What is the free charge q before the slab is inserted? _______________ C (d) What is the free charge q after the slab is inserted? ________________C (e) What is the magnitude of the electric field in the space between the plates and dielectric? _______________V/m (f) What is the magnitude of the electric field in the dielectric itself? _______________ V/m (g) With the slab in place, what is the potential difference across the plates? ________________ V (h) How much external work is involved in the process of inserting the slab? ________________ JExplanation / Answer
A) (C= epsilon_{o}*A/d =112pF)
B) (C_{new}= [epsilon_{o}*A/d_{1}]+[kepsilon_{o}*A/d_{2}] =142pF)
C) (Q=C*V=1.57 * 10^{-8} C)
D) (Q_{new}=1.57 * 10^{-8} C) bec battery has been removed before dielectric is inserted.
E) (E=V/d_{1}=9328.2 V/m)
F) (E_{dielectric} = V_{dielectric}/d_{2} =9347.5 V/m)
G) (V_{plate} = Q/C_{new} = 110.56V)
H) (work done = Delta U = 2.30*10^{-7} J)