A parallel plate capacitor consists of two plates of surface area (A) spearated by a distance (d.) This capacitor is in turn connected in series to a DC supply of voltage V volts. A slab of material with permitivity k and thickness d/2 is placed between the plates as shown (SEE PROBLEM ATTACHMENT). We can divide the dielctric between the plates into two regions: region I of thickness d/2 and with dielectric constant of free space and another of region II and dielectric constant k and thickness d/2
PART A: What is the electric field in region I?
PART B: What is the electric field in region II?
PART C: What is the equivalent capacitance of the capacitor?© BrainMass Inc. brainmass.com October 25, 2018, 3:41 am ad1c9bdddf
PART A: Solution.
Capacitance is given by
C = Epsilon*A/d (1)
where A is the surface area of a plate, d is distance between plates for a || plate capacitor
Also we can use the relation Q = CV (2)
V = Q/C (3)
for the voltage developed across the plates of a capacitor in terms of Q the charge stored and C the capacitance as given by (1)
Substitution of (1) in (2) we get
V = Qd/Epsilon*A (4)
Electric field can be expressed as potential gradient with distance and is given as (5)
E = V/d (5)
Thus if we divide (4) by d we get a general expression for the ...
A hybrid parallel plate capacitor is presented which consists of two differing dielectrics between the || plates. The problems revolve around determining the electric field strengths in the two separate regions based on given paraments such as the dielectric constants for each region and the thickness of each region. The problem then goes on to find an answer to determine the effective and equivalent capactiance of the described device. A step by step approach, using linear algebra techniques, is shown, showing how to derive such quantities.
Two problems on capacitors
1. Two parallel plates of area 100cm^2 are given charges of equal magnitudes 8.9X10^-7 but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4X10^6 V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
2. You are asked to construct a capacitor having a capacitance near 1 nF and a breakdown potential in excess of 10000V. You think of using the sides of a tall Pyrex drinking glass as a dielectric, lining the inside and outside curved surfaces with aluminum foil to act as the plates. The glass is 15cm tall with an inner radius of 3.6cm and an outer radius of 3.8cm. What are the (a) capacitance and (b) breakdown potential of this capacitor?View Full Posting Details