CAPACITANCE
&
DIELECTRICS
Group: 4
CAPACITANCE & DIELECT RICS
LEARNING
OUTCOMES
• Calculate the capacitance of a given
capacitor;
• Calculate the equivalent capacitance of
capacitors connected in series or parallel;
• Calculate the amount of energy stored in a
capacitor;
• Explain the role of a dielectric in a
capacitor;
• Apply Gauss’s law to dielectrics to solve for
the capacitance.
15/06/2023
2
CAPACITANCE & DIELECTRICS
the capacitance of a given capacitor
3
the capacitance of a given capacitor
The calculation of capacitance for a given capacitor involves determining the amount of electrical charge
the capacitor can store per unit voltage applied across its terminals. Capacitance is measured in farads
(F) and is influenced by the physical characteristics of the capacitor, the dielectric material between its
plates, and the geometric arrangement of its components.
The capacitance equation for a parallel plate capacitor, which is a common type, is C = ε₀ * (A / d),
where C represents the capacitance, ε₀ is the permittivity of free space, A denotes the surface area of
the capacitor plates, and d is the distance between the plates. A larger plate area or smaller plate
separation results in a higher capacitance value.
9/3/20XX
PRESENTAT ION T IT LE
4
When a dielectric material is present between the capacitor plates, the equation becomes C = κ * ε₀ *
(A / d), where κ is the dielectric constant of the material. Different dielectric materials have varying
dielectric constants, which significantly impact the capacitance of a capacitor.
For more complex capacitor designs, additional formulas or numerical methods may be required to
calculate the capacitance. In summary, the calculation of capacitance involves considering the
physical properties, dielectric material, and geometric arrangement of the capacitor to determine its
capacitance value.
9/3/20XX
PRESENTAT ION T IT LE
5
CAPACITANCE & DIELECTRICS
the equivalent capacitance of capacitors
connected in series or parallel
6
the equivalent capacitance of capacitors connected in series or
parallel
When capacitors are connected in series or parallel, the equivalent capacitance of the combination can
be calculated.
Capacitors in Series: When capacitors are connected in series, the total or equivalent
capacitance (Ceq) is given by the reciprocal of the sum of the reciprocals of individual
capacitances (C1, C2, C3, etc.): 1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...
In series connection, the total capacitance is always smaller than the smallest individual capacitance.
9/3/20XX
PRESENTAT ION T IT LE
7
Capacitors in Parallel: When capacitors are connected in parallel, the total or equivalent capacitance
(Ceq) is the sum of individual capacitances: Ceq = C1 + C2 + C3 + ...
In parallel connection, the total capacitance is always larger than any individual capacitance.
These formulas allows to calculate the overall capacitance when capacitors are connected in series or
parallel. It is important to note that these formulas only apply when the capacitors are connected
directly to each other, without any other components or elements in the circuit between them.
9/3/20XX
PRESENTAT ION T IT LE
8
the amount of energy
stored in a capacitor
When a capacitor is connected to a circuit, the
positive pole of the voltage source begins to push the
electrons from the plate to which it is connected.
These pushed electrons gather in the other plate of
the capacitor, causing excess electrons to be stored
in plates
E = 1/2 * C * V^2
E is the energy stored in joules, C is the
capacitance of the capacitor in farads, and V is
the voltage across the capacitor in volts
9
PRESENTAT ION T IT LE
The energy stored in capacitor is the product of the
charge on the capacitor and the potential difference
between its plates divided by two.
Calculate the energy stored in a capacitor with a capacitance of 60 F and a
voltage of 100 V.
Solution:
A capacitor with a capacitance of 60 F is charged to a voltage of 100 V. The
capacitor's stored energy can be calculated as follows
E = 1/2 x C x V^2
E = 1/2 x 60 x 100^2
= 300 x10^3 J
9/3/20XX
PRESENTAT ION T IT LE
10
CAPACITANCE & DIELECTRICS
A dielectric is a non-conductive material that is
placed between the plates of a capacitor.
the role of a dielectric
in a capacitor
Its primary role is to increase the capacitance of
the capacitor by reducing the electric field
between the plates.
The dielectric material is typically an insulator,
such as ceramic, plastic, glass, or paper, which
prevents the flow of electric current.
When a dielectric is placed between the plates of
a capacitor, it decreases the electric field intensity
within the capacitor by polarizing the molecules
within the dielectric.
7
This polarization produces an opposing electric field that cancels out the electric
field between the plates, lowering the effective voltage across the capacitor.
Capacitors can store more charge for a given voltage by using a dielectric, making
them beneficial in a variety of electronic applications.
Other advantages of dielectrics include enhanced insulating between plates,
protection against short circuits, and less reactivity to temperature fluctuations .
9/3/20XX
PRESENTAT ION T IT LE
12
Gauss’s law to
dielectrics to solve for
the capacitance
Gauss's law states that the electric flux
passing through a closed surface is equal to
the total charge enclosed divided by the
electric constant (ε₀). For a parallel-plate
capacitor with a dielectric material between the
plates, we can consider a Gaussian surface in
the shape of a cylinder placed symmetrically
between the plates.
By assuming the electric field inside the
dielectric to be E and the charge density on
the plates to be σ, we can simplify the
equation to
(E/εᵣ)A - (E/εᵣ)A = (σA) / ε₀.
13
CAPACITANCE & DIELECTRICS
Applying Gauss's law to dielectrics allows us
to determine the capacitance of a capacitor.
Through further analysis, we find that the charge density on the plates is zero for a dielectric,
meaning the electric field inside the dielectric is solely due to polarization. By calculating the
electric field inside the dielectric and the potential difference (V) between the plates, we can
determine that the capacitance (C) of the parallel-plate capacitor with a dielectric is zero. This
simplified explanation demonstrates how Gauss's law can be used to solve for capacitance in
dielectric systems
9/3/20XX
PRESENTAT ION T IT LE
14
CAPACITANCE & DIELECTRICS
9/3/20XX
15
THANK YOU
Presenter name:
SIBAL, ACEYORK N.
MANZANERO, ANDREI LIOYD II
ADOLFO, ALDRIN JOHN
EVANGALISTA, DAN ABRAHAM
BALITA, KURT ROMIEL