To test for a negative sign, the initial charge
on the electroscope must be positive.
If the material place near the cap causes a
collapse of the leaf, then this is clear
indication of the presence of a negative
charge on the material. Example if a glass
rod is rubbed with silk and brought near a
brass cap, there is decrease in the leaf
divergence.
Note that: A decrease in divergence
(attraction is not a sure test) because of the
following reasons.
(i) Attraction also occurs on uncharged
body due to induction
(ii) Attraction occurs between opposite sign
Therefore, a decrease in divergence however
does not necessarily mean that a charge of
opposite kind is near a charged electroscope.
Increase in divergence is therefore, a sure
test.
Identifying the insulating properties of
materials
An electroscope that is positively charged
can be used to test for the insulating
properties of materials.
If the material that is placed near the cap of
an electroscope is a conductor, then the
metal leaf collapses.
If the material being tested is an insulator,
the leaf retains its charge and remains raised.
Detecting the presence of charge on a
body
When a charge is induced on the
electroscope by a charged body, the leaf
diverges. When the charged body is
removed, the leaf collapses indicating that
the induced charge on the electroscope is
temporary and it is due to the charged body.
Example 01
An insulating rod A is charged and brought
near an electroscope as shown in the figure
below.
With rod A is still near the cap of the
electroscope, the cap is briefly touched and
rod A is removed. It is observed that the leaf
of the electroscope remains diverged as
shown below.
Is the charge on the electroscope now the
same as the charge that was on the rod A or
the opposite charge? Explain your answer.
Answer
The charge on the electroscope is now the
same as the charge that was on the rod A.
This is because, the leaf remain diverged. If
the charges are opposite the leaf should
collapse indicating attraction.
Example 02
A sharp needle was brought close to the cap
of a charged leaf electroscope. Explain why
the leaf collapsed.
Answer
The leaf collapsed because the sharp needle
has charges opposite to those of the gold –
leaf electroscope.
Example 03
(a) Draw the sketch of a large and well
labeled gold – leaf electroscope
Answer
(b) How is the electroscope used for
testing the type of charges?
Answer
Testing for a negative charge, the
electroscope must be positively
charged. The leaf collapses if the
negatively charged is brought nearby
the cap of the electroscope.
Testing for a positive charge, the
electroscope must be negatively
charged. The leaf collapses if the
positively charged object is brought
near the cap of the electroscope.
POTENTIAL DIFFERENCE
Potential difference (p.d) is the work done
to move a charge from one point to another
when the points are at different potentials.
The SI – Unit of potential difference is
Volt(V)
A volt is one joule of work per electric
charge of one coulomb.
It should be understood that, leaf divergence
in a gold leaf electroscope occurs because
there is a potential difference between the
leaf and the cap.
Electrons move from point of low potential
to points of high potential as shown on the
figure below until both are at the same
potential.
1.30 CAPACITORS
A capacitor is a device used to store electric
charge.
It consists of two electrically conductive
plates, separated by insulator materials
(dielectric). The shape of a capacitor can be
square, circular or cylindrical.
Dielectric is an insulating medium used
between the plates of a capacitor. Example
ceramic, plastic, air, paper mica or liquid
gel.
The symbol for capacitor is shown below
Action of a capacitor
When power source is connected across the
plates, one plate is charged negatively, and
the other positively. Charges continue to
accumulate on the plates with time until the
capacitor is fully charged.
A fully charged capacitor has a positive
charge on one plate and an equal amount of
negative charge on another. A voltmeter
connected between the capacitor plates
measures the potential difference between
the plates.
Applications of capacitors
(i) Are used for energy store
(ii) Are used in photoflash units
(iii)Are used for tuning radios
(iv) Are used for power conditioning
(v) Are used fir signal processing
Capacitance is the measure of the amount
of charge a capacitor can hold for a given
potential difference.
capacitance=
Amount of charge stored
potential difference
C=
Q
V
The SI – unit of capacitance is Farad, F.
Farad is the capacitance of a capacitor
when a charge of one coulomb changes its
potential difference by one volt.
Other units are millifarad (mF), microfarad
(μF), Picofarad (pF), nanofarad (nF)
1mF= 10
-3
F
1μF= 10
-6
F
1nF= 10
-9
F
1pF= 10
-12
F
To find the quantity of charge stored in the
capacitor, Q of capacitance, C when the
potential difference, V is applied at the
capacitor. We use the formula.
Q= C×V
Example 01
A capacitor of capacitance 200μF is being
charged. The potential difference (p.d)
between its plates is 10V. How much charge
will accumulate on the plates during the
period of charging?
Solution
Step I: To convert the capacitance in Farad
(F)
Given
C= 200μF
=200× 10
-6
F
=2× 10
-4
F
Step II: To find the quantity of charges, Q
Q= C×V
= 2× 10
-4
F×10V
= 2× 10
-3
C
= 2 mC
Therefore, one plate has a charge of +2 mC
and the other has -2 mC
Example 02
A capacitor with a capacitance of 50 pF is
charged to 30V. What is the charge on its
plates?
Solution
Step I: To convert the capacitance in Farad
(F)
Given
C= 50pF
=50× 10
-12
F
=5× 10
-11
F
Step II: To find the quantity of charges, Q
Q= C×V
= 5× 10
-11
F×30V
= 1.5× 10
-9
C
Therefore, one plate has a charge of -1.5×
10
-9
C and the other +1.5× 10
-9
C
Example 03
If a cell has a voltage of 1.5V, calculate the
capacitance of the capacitor when the charge
is 90 coulombs.
Solution
Capacitance=
amountofchargeonplates
p.d
between
the
plates
Capacitance=
90 C
1.5V
Capacitance= 60 F
The capacitance of the capacitor is 60 Farad.
Example 04
A 3μF capacitor has a p.d of 12V.
Determine the total charge on the capacitor.
Solution
Step I: To convert the capacitance in Farad
(F)
Given
C= 3μF
=3× 10
-6
F
Step II: To find the quantity of charges, Q
Q= C×V
= 3× 10
-6
F×12V
= 36× 10
-6
C
= 3.6× 10
-5
C
Therefore, one plate has a charge of -3.6×
10
-5
C and the other +3.6× 10
-5
C
Example 05
Briefly explain the following
(a) Capacitor
Answer
A capacitor is a device used to store
charges
(b) Capacitance
Answer
Capacitance is the ratio of charge on
the capacitor to the potential difference
across the plates of the capacitor.
TYPES OF CAPACITORS
Based on the dielectric materials used to
make the capacitor, there are different types
of capacitors as follows
Paper or plastic capacitors
