The capacitance of a capacitor with a dielectric of air is 1200 pF. When a dielectric is inserted between the plates, the capacitance increases to 0.006 ìF. What is the value of e.
ans= 1200/0.006=
To find the value of e, we need to use the equation that relates the capacitance of a capacitor to the dielectric constant:
C = (e * A) / d
Where:
C = capacitance of the capacitor
e = dielectric constant
A = area of the capacitor's plates
d = distance between the plates
In this case, we are given two different capacitance values:
C1 = 1200 pF = 1200 * 10^-12 F
C2 = 0.006 uF = 0.006 * 10^-6 F
Since the area and distance between the plates remain the same, we can set up the following equation:
(C2 * d) / A = (C1 * d) / A * e
Simplifying the equation, we get:
e = (C2 * d) / (C1 * d)
Plugging in the given values, we have:
e = (0.006 * 10^-6 F) / (1200 * 10^-12 F)
Simplifying further:
e = (0.006 / 1200) * 10^6
So, the value of e is:
e = 5 * 10^3