Question

(Chapter 18-Homework Practice Problem 18.10 k 5 of 22 Now let's examine the effocts of placing a diseletic fi between the plates of a capacitor The plates of an ar tled parallel-plate capacitor each have area 2.00 x 10 em2 and aro 1.00 cm apart. The capactor is connected to a power supply and charged to a potential dfference of Vo3.00 kV. The capacitor is then deconnected from the power supply without any charge being lost from its plates After the capacitor has been disconnected, a sheet of insulating plastic material i Part C Practice Problem: Find the electric field between the plates before the dielectric is inserted Express your answer in volts per meter to three significant figures nserted between the plates, completely filling the space between them. When this is done, the petontial difforence betwoon the plates decreases to 1 00 kV, while the charge remains constant (a) What is the capacitance of the capactor before and after the dielectnc is inserted? V/m What is the dielectic constant of the plastie? (c) What is the electric field between the plates before and aher the Selectric is inserid Figure of 2 Part D- Practice Problem Find the electhic field between the plates after the dielectric is nserted Express your answer in volts per meter to three signilicent figures d 100 C 8

Answer 1

here, we have

C = 8.85e-12 * 0.2 / 0.010 = 1.77e-10 or 177 pF

original charge

Q = CV = 1.77E-10 * 3e3 = 5.31e-7 = 0.531 uC

After dielectric is inserted, V = 1 kV

so, C = 5.31e-7 / 1e3 = 5.31e-10 C = 531 pF

so, electric field before insertion = 3e3 / 0.01 = 3e5 V/m

electric field after insertion = 1e3 / 0.01 = 1e5 V/m

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Now, use the same formula

E = V/d

where V is the new potential difference given which you did not provide

you need to use same formulas again.

just replace V = 3 kV with new value given in practice problem ( both for before and after insertion of dielectric)

Let me know if you have any doubt.