A capacitator with air between its plates is charged to 100V and then disconnected from the battery. When a piece of glass is placed between the plates, the voltage across the capacitator drops to 25V. What is the dielectric constant of the glass? (assume the glass completely fills the space between the plates.) I am given the equation: C=k(Esubzero)(Area/distance). Thank you for any help!
2007-02-28 14:09:28 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Q = CV so V = Q/C
Because V goes down by 4, C must have gone up by 4. Q remains fixed because you didn't add or subtract any charge.
Since C is proportional to the dielectric constant, the dielectric constant mus have gone up by 4. Since the dielectric constant of air is about 1.0, the glass must have a dielectric constant of 4.0
2007-02-28 14:18:44 · answer #1 · answered by hello 6 · 0⤊ 0⤋
the charge created with the air dielectric
q=εoεkA/dV ==> q=1.00054 A/d 100
The q doesn't change
q=x A/d 25
1.00054 *100 = x *25 ==> x=4.022
2007-02-28 22:27:29 · answer #2 · answered by Rob M 4 · 0⤊ 0⤋