Two parallel plates are .005m apart and are each 2m2 in area. The plate are in v

Question

Two parallel plates are .005m apart and are each 2m2 in area. The
plate are in vacuum and an electric potential difference of 10,000V is
applied across them.
A) Find the capacitance, the charge on each plate, the electric field intensity
in the space between the plates, and the stored energy.
B) If a dielectric material with dielectric constant ?=80.4 is inserted into the
gap between the plates, with the electric potential remaining the same, find
the new capacitance, charge on each plate, electric field intensity, and
energy stored.

Explanation / Answer

A) for plate capacitor

C=0*A/d=0*2/0.005=3.54e-9 F=3.54 nF

Q=V*C=3.54e-5 C

E=V/d=10000/0.005=2e6 V/m

B)K=80.4

C=k*0*A/d=0*80.1*2/0.005=2.87e-7 F

Q=V*C=2.87e-3 C

E=V/d=2e6 V/m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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