A parallel plate capacitor has square plates with sides of length 19 cm. The dis

Question

A parallel plate capacitor has square plates with sides of length 19 cm. The distance between the plates is 1 mm. The plates are charged up to 30 volts.

A.) What is the electric field between the plates? -- I already have this, it's 30,000 N/C

B.) What is the amount of charge on each plate? (unit for answer: C)

C.) What is the capacitance? (in F)

D.) What is the energy stored by the capacitor? (in nJ)

E.) The space between the plates is filled with a dielectric which has a dielectric constant of 20, and the new capacitor is charged up to the same voltage as before. How much energy is stored in the capacitor now? (in nJ)

Thanks in advance for your help!!!

Explanation / Answer

here,

a = 0.19 m

distance between the plates , d = 0.001 m

V = 30 V

C)

capacitance , C = area * epsilon0 /d

C = 0.19^2 * 8.85 * 10^-12 /( 0.001)

C = 3.19 * 10^-10 F

C = 3.19 * 10^-4 uF

B)

amount of charge on plates , Q = C * V

Q = 3.19 * 10^-10 * 30

Q = 9.57 * 10^-9 C

D)

energy stored , E = 0.5 * C * V^2

E = 0.5 * 3.19 * 10^-10 * 30^2

E = 1.44 * 10^-7 J

E = 143.55 nJ

the energy storedd in the capacitor is 143.55 nJ

E)

the energy stored in the capacitor now , E = 0.5 * k * C * V^2

E = 0.5 * 20 * 3.19 * 10^-10 * 30^2

E = 2871 nJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.