Initially the space between the plates of each capacitor is filled with air. Whi
ID: 1925589 • Letter: I
Question
Initially the space between the plates of each capacitor is filled with air. Which of the following changes will double the total amount of charge stored on both capacitors with the same applied potential difference?
a) Fill the space between the plates of both C1 and C2 with Teflon (dialectric constant of 2)
b) FIll the space between the plates of both C1 and C2 with glass (dielectric constant of 4)
c) Fill the space between the plates of C1 with glass (dielectric constant of 4) and leave C2 as is.
d) Fill the space between the plates of C1 with Teflon (dielectric constant of 2) and leave C2 as is.
**Please provide brief explanation. Thank you!
Explanation / Answer
We know that Capacitance, C = ro * A/ d
C is the capacitance;
A is the area of overlap of the two plates;
r is the relative static permittivity (sometimes called the dielectric constant) of the material between the plates (for a vacuum, r = 1);
0 is the electric constant (0 8.854×1012 F m–1); and
d is the separation between the plates.
And we also know, Charge, Q = C* V
=> Q is directly proportional to r
A.) Teflon with r = 2 is introduced, hence C will be doubled and so is the charge
B.) Filled with r = 4, hence charged will be quadrapled not doubled!
C.) Only one plate is changed with r = 4. hence, charge q1 will be 4*C*V and the second q2 will be same. C*V. New q1 + q2 = 5*C*V = 2.5 * (C*V + C*V) = 2.5 * (Old q1 + old q2)
D.) Only one plate is changed with r = 2. hence, charge q1 will be 2*C*V and the second q2 will be same. C*V. New q1 + q2 = 3*C*V = 2.5 * (C*V + C*V) = 1.5 * (Old q1 + old q2)
Hence, the answer is "A"