Capacitors

Question 1

What is a capacitor

Answer 1

A device that can store charge

Question 2

If one charged plate of the capacitor stores a charge Q, what charge is stored by the other plate

Answer 2

-Q

One is Q other is -Q

One plate gains an amount of charge, in turn causing the other plate to lose the same amount of charge

Question 3

Definition of capacitance

Answer 3

Charge stored per unit p.d.

For a capacitor storing charge Q at a p.d V, C = Q/V

Question 4

Uses of capacitors

Answer 4

Smoothing circuits (w unwanted variations in voltage)

Back up power supplies (When mains is interrupted)

Filter circuits (Removing unwanted frequencies)

Tuning circuits (Used to select radio stations/TV channels)

Question 5

Why a charged capacitor stores energy

Answer 5

Electrons forced onto a plate, and taken off another - Energy stored in capacitor as electric potential

Question 6

Deriving energy stored in a capacitor

Answer 6

Capacitance C, when charge increases from q to q + Δq. Energy stored ΔE equals work done to force extra charge onto plates = vΔq where v is average pd across capacitor in this step. Total energy stored given by adding up strips (form a triangle) E=1/2QV

E = 1/2 CV² or Q²/2C

Question 7

Energy transferred to charging/dissipated during charging of a capacitor

Answer 7

Battery forces charge Q through pd V round the circuit, transfers energy QV to the circuit - 50% of energy stored in capacitor (1/2QV) while other 50% wasted due to resistance of circuit as it’s transferred to surroundings when charge flows in circuit.

Question 8

Energy stored in a thundercloud

Answer 8

Imagine thundercloud and earth as a pair of parallel charged plates. Electric field of field strength V/d exists between them.

Energy stored = 1/2QV = 1/2QEd. If thundercloud rises to new height d’, new energy stored = 1/2QEd’ (field strength unchanged as it depends on charge/area). Increase = 1/2QEΔd.

Energy increases as work is done by force of the wind to overcome attraction between thundercloud and earth. Insulating property of air breaks down in electric fields stronger than 300kV/m

Question 9

Showing how current,charge and p.d decrease exponentially. (Consider resistor connected across discharging capacitor)

Answer 9

Current through resistor at a point = Pd/R
Q/CR, in a time t, ΔQ=-IΔt so ΔQ/Δt = -Q/CR

ΔQ/Q = -Δt/CR

Tells us that the fractional drop in charge is the same for every interval - 0.9 - 0.81 - 0.729 - In theory charge never becomes zero

also ΔQ/Δt = -Q/CR

Solution is Q=Q₀e^(t/RC)

Question 10

What is the time constant

Answer 10

RC, at time t=RC, charge falls to 0.37(1/e)(37%) to its original value

Units are seconds as exponents must always be dimensionless –
(-t/RC)

5RC used more as it’s time to discharge over 99% of charge

Question 11

Why oscilloscope is used in calculating time constant experiment

Answer 11

Has a high resistance so discharge current only passes through fixed resistor

Question 12

Charging capacitor derivation

Answer 12

At any instance, source pd = resistor pd + capacitor pd = IR + Q/C

Initial current I₀=V₀/R, assuming capacitor is initially uncharged. Time t after charging I= I₀…

Get V=V₀e^(-t/RC) + Q/C

Q/C = V₀(1-e^(-t/RC))
As V=Q/C can sub in

t=0, Q,V=0
t tends to inf - Q,V - Q₀V₀

Question 13

Opposites in charging/discharging

Answer 13

When charging, Q and V increase wrt time, opposite when discharging. However current always decreases wrt time, as in charging case - zero potential diff between source and capacitor, and in discharging case - rate of change of charge is negative

Question 14

Permittivity/rel

Relative also called dielectric constant

Answer 14

Permittivity - how difficult it is to generate an electric field in something, relative is ratio of material to free space εr = εm/ε0

also Q/Q0 where Q is charge stored by plate when space is completely filled w material, Q0 when space is empty

also C/C0

Question 15

How dielectrics increase capacitance

Molecules can be polar/non polar

Answer 15

When no charge stored between plates, no electric field, molecules point in random directions.

When charge, electric field created across capacitor. Negative ends of molecules attracted to positive plate and vice versa, aligning themselves w electric field. Molecules have their own electric field, which opposes the applied electric field of capacitor. Larger ε larger opposition. Reduces overall electric field, reducing potential difference needed to fully charge, increasing capacitance. (also can use eqs)

Also can say that dielectric molecules cause surface facing plate to be opposite - e.g. negative side will be on positive plate, attracting more “positive charge” from battery

Question 16

Why ionic crystals dissolve easily in water

Answer 16

Water has a high dielectric constant

Question 17

Capacitance eq relating area and d

Answer 17

C=Aεoεr/d where A is the surface area of each plate and d is the distance between them

Question 18

Capacitor lamp caveat

Answer 18

A charged capacitor will discharge through a lamp, giving a brief flash of light given that it has been charged to operating pd

Question 19

Why capacitance is inversely proportional to d

Answer 19

This is because the strength of the electric field is the voltage divided by distance. For fixed voltage, the field is inversely proportional to this separation. A stronger field allows more charge, thus, higher capacitance

For area, more area between the plates means more charge can be stored in the capacitor

Question 20

Misteak i made w simplifying energy equation

Answer 20

dE = 1/2 Cdv^2 but dv is V2^2-V1^2
not (V2-V1)^2