ELECTRIC POTENTIAL ALI ALZAHRANI

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ELECTRIC POTENTIAL
ALI ALZAHRANI
ELECTRIC POTENTIAL

Electric potential energy is the energy needed to change the spatial configuration of charged
particles from initial to final destinations.
βˆ†π‘ˆ = π‘ˆπ‘“ − π‘ˆπ‘– = −π‘Š

We used to take the energy at infinity is zero, therefore, the energy required to move any
charged particle to any position is
π‘ˆ = −π‘Š

Electric potential ‘potential’ is defined as the potential energy per unit charge, such as
𝑉=
π‘ˆ
π‘ž
ELECTRIC POTENTIAL

The potential difference between any two points is defined as
βˆ†π‘‰ = 𝑉𝑓 − 𝑉𝑖 =

π‘ˆπ‘“ π‘ˆπ‘– βˆ†π‘ˆ
π‘Š
− =
=−
π‘ž
π‘ž
π‘ž
π‘ž
The SI unit of potential is ‘volt (V)’ which is also J/C. However, the potential energy is
measured by ‘electron-volt (eV)’ which is 1.6 × 10−19 J.
1 eV = 1.6× 10−19 C. (1 J/C) = 1.6× 10−19 J

Equipotential surfaces are those where each point on them have the same electric potential
or the potential difference between any two points is zero. Therefore no net work should be
done on a charged particle to move between any two points on that surface.
EQUIPOTENTIAL SURFACE - EXAMPLES
(a) A single positive charge
(c) Two positive charges
(b) An electric diploe
(d) Uniform electric field
ELECTRIC POTENTIAL AND FIELD

The electric potential is defined in terms of electric field as
𝑓
βˆ†π‘‰ = −
𝐸. 𝑑𝑠
𝑖
From this we note that 1 V = 1 N.m/C or 1 N/C = 1 V/m. If the initial potential is zero, the
above equation becomes
𝑓
𝑉=−
𝐸. 𝑑𝑠
𝑖
NOTE: It is important to note that the potential is a scalar quantity NOT a vector.
ELECTRIC POTENTIAL DUE TO A POINT CHARGE

The electric potential of a point charge π‘ž at any distance π‘Ÿ can be evaluated from
𝑓
𝑉=−
𝐸. 𝑑𝑠
𝑖
Having known that the electric field due to any point charge is 𝐸 =
π‘Ÿ
𝑉=−
π‘˜π‘ž
π‘˜π‘ž
π‘‘π‘Ÿ
=
2
π‘Ÿ
∞ π‘Ÿ
NOTE: Positive charges produce positive potential and vice versa.
π‘˜π‘ž
π‘Ÿ2
we get
ELECTRIC POTENTIAL DUE TO A GROUP OF POINT CHARGES

The electric potential of a collection of point charge π‘žπ‘– at any distance π‘Ÿπ‘– is the
summation of all potentials of each individual point charge
𝑛
𝑉=
𝑛
𝑉𝑖 = π‘˜
𝑖=1
𝑖=1
π‘žπ‘–
π‘Ÿπ‘–

The electric potential due to an electric dipole 𝑝 at any point at distance π‘Ÿ and makes an
angle πœƒ with the axes of the dipole is defined as
π‘˜ 𝑝 cos πœƒ
π‘Ÿ2
NOTE: The angle πœƒ is measured from the direction of the dipole.
𝑉=
ELECTRIC POTENTIAL DUE TO A CONDUCTING SPHERE

The electric potential of a conducting sphere of radius 𝑅 and charge π‘ž at any distance π‘Ÿ outside the
sphere, where the electric field (outside) is 𝐸 =
π‘˜π‘ž
π‘Ÿ2
, is
π‘Ÿ
𝑉=−

π‘˜π‘ž
π‘˜π‘ž
π‘‘π‘Ÿ
=
2
π‘Ÿ
∞ π‘Ÿ
The electric potential of an insulating sphere of radius 𝑅 and charge π‘ž at any distance π‘Ÿ inside the
sphere, where the electric field (outside) is 𝐸 =
𝑅
𝑉=−
∞
π‘˜π‘ž
π‘Ÿ2
π‘˜π‘ž
π‘‘π‘Ÿ −
π‘Ÿ2
and (inside) 𝐸 = 0, is
π‘Ÿ
0 π‘‘π‘Ÿ =
𝑅
π‘˜π‘ž
𝑅
Therefore the electric potential at any point of the isolated conductor is invariant, leading to that no
extra work is needed to carry a charge from surface to any point within the conductor, see example 7.
ELECTRIC POTENTIAL DUE TO AN INSULATING SPHERE

The electric potential of an insulating sphere of radius 𝑅 and charge π‘ž at any distance π‘Ÿ
outside the sphere, where the electric field (outside) is 𝐸 =
π‘˜π‘ž
π‘Ÿ2
, is
π‘Ÿ
π‘˜π‘ž
π‘˜π‘ž
π‘‘π‘Ÿ
=
2
π‘Ÿ
∞ π‘Ÿ
𝑉=−

The electric potential of a conducting sphere or radius 𝑅 and charge π‘ž at any distance π‘Ÿ
inside the sphere, where the electric field (outside) is 𝐸 =
𝑅
𝑉=−
∞
π‘˜π‘ž
π‘‘π‘Ÿ −
π‘Ÿ2
π‘Ÿ
𝑅
π‘˜π‘ž
π‘Ÿ2
and (inside) 𝐸 =
π‘˜π‘ž π‘Ÿ
π‘˜π‘ž
π‘Ÿ2 − 𝑅2
π‘‘π‘Ÿ =
1−
𝑅3
𝑅
2𝑅 2
π‘˜π‘ž π‘Ÿ
,
𝑅3
is
POTENTIAL ENERGY OF A SYSTEM OF POINT CHARGES

The electric potential energy of a system of point charges π‘žπ‘– separated by distances π‘Ÿπ‘–
is the summation of all potential energy of these point charges
𝑛
π‘ˆ=
𝑛
π‘ˆπ‘–π‘— = π‘˜
𝑖,𝑗=1
𝑖,𝑗=1
where π‘Ÿπ‘–π‘— is the distance between charges π‘žπ‘– and π‘žπ‘—
π‘žπ‘– π‘žπ‘—
π‘Ÿπ‘–π‘—
E DERIVED FROM V

The electric field can be evaluated from the expression of the potential such as
𝐸=−
∂𝑉
∂𝑠
In our three-dimensional space, the above definition can be rewritten as
𝐸π‘₯ = −
πœ•π‘‰
πœ•π‘₯
𝐸𝑦 = −
πœ•π‘‰
πœ•π‘¦
Therefore the electric field vector is
𝐸=−
πœ•π‘‰
πœ•π‘‰
πœ•π‘‰
𝑖−
𝑗−
π‘˜
πœ•π‘₯
πœ•π‘¦
πœ•π‘§
𝐸𝑧 = −
πœ•π‘‰
πœ•π‘§
1. Calculate the electric potential at a distance 5 cm from a point charge 2.5 nC.
Solution
The magnitude of the electric potential at any distance from a point charge is given by
𝑉=
π‘˜π‘ž
π‘Ÿ
Hence the electric potential at 5 cm from the charge is
𝑉=
π‘˜π‘ž
π‘Ÿ
=
9×109 ×2.5×10−9
0.05
= 450 V
2. A point charge produces an electric field of 180 N/C at 2 cm. Calculate the electric potential at 4 cm from
the charge.
Solution
The magnitude of the electric field due to a point charge is given by
𝐸=
π‘˜ π‘ž
π‘Ÿ2
Therefore to find the electric potential at any point, we have to determine the charge, firstly.
π‘ž=
𝐸 π‘Ÿ2
π‘˜
=
180×0.022
9×109
= 8 × 10−12 C
Hence the electric potential at 4 cm from the charge is
𝑉=
π‘˜ π‘ž 9 × 109 × 8 × 10−12
=
= 1.8 V
π‘Ÿ
0.04
3. As shown in the arrangement below, calculate the electric potential at point A. Calculate
the work needed to bring a charge Q=6 nC from infinity to point A.
Solution
A

2m
3 µC
3m
-4 µC
The electric potential of a collection of points is
𝑉 = 𝑉1 + 𝑉2 =
π‘˜ π‘ž1 π‘˜ π‘ž2
π‘ž1 π‘ž2
3 × 10−6 −4 × 10−6
+
=π‘˜
+
= 9 × 109 ×
+
= 6300 V
π‘Ÿ1
π‘Ÿ2
π‘Ÿ1 π‘Ÿ2
2
5
Therefore the work needed to bring a charge from infinity to this point is
π‘Š = −𝑄𝑉 = −6 × 10−9 × 6300 = −3.78 × 10−5 J = −2.36 × 1014 eV
4. Four charges 1, 2, 3 and -4 µC are located at the corners of a square of side 3 m as shown in
the figure. Calculate the electric potential energy of the system.
Solution
q2=2 µC
q3=3 µC
q1=1 µC
q4=-4 µC
The potential energy is
π‘˜π‘ž1 π‘ž2 π‘˜π‘ž2 π‘ž3 π‘˜π‘ž3 π‘ž4 π‘˜π‘ž4 π‘ž1 π‘˜
9 × 109
π‘ˆ=
+
+
+
= π‘ž1 π‘ž2 + π‘ž2 π‘ž3 + π‘ž3 π‘ž4 + π‘ž4 π‘ž1 =
2 + 6 − 12 − 4 × 10−12 = −24 mJ
π‘Ÿ12
π‘Ÿ23
π‘Ÿ34
π‘Ÿ41
π‘Ÿ
3
5. Two equal and opposite charges 6.0 µC and -6.0 µC are separated by a distance of 2 cm.
What is the electric potential at a point 30 cm away from the axis of the dipole?
Solution
The electric potential due to a dipole is
𝑉=
π‘˜ 𝑝 cos πœƒ
π‘Ÿ2
However along the axis of the dipole the angle is zero, (cos πœƒ = 1), therefore
𝑉=
π‘˜ 𝑝 π‘˜ π‘žπ‘‘ 9 × 109 × 6 × 10−6 × 0.02
= 2 =
= 12 kV
π‘Ÿ2
π‘Ÿ
0.32
6. A metallic sphere of radius 10 cm has a charge of 5.0 µC. What is the electric potential at (i)
5 cm and (ii) 15 cm from its center?
Solution
(i) The electric potential at any point inside a conducting sphere is
𝑉=
π‘˜ π‘ž 9 × 109 × 5 × 10−6
=
= 450000 V
𝑅
0.1
(ii) The electric potential at any point outside a conducting sphere is
π‘˜ π‘ž 9 × 109 × 5 × 10−6
𝑉=
=
= 300000 V
π‘Ÿ
0.15
7. A conducting sphere of radius 10 cm has a charge of 5.0 µC. Calculate the work required to
carry a charge 2.0 µC from the surface to the center of the sphere.
Solution
We know that the electric potential at any point inside a conducting sphere is
𝑉=
π‘˜π‘ž
𝑅
𝑉=
π‘˜π‘ž
𝑅
Also the electric potential at the surface is
Therefore the potential difference is zero, leading the work is zero, too, according to π‘Š = −𝑄 βˆ†π‘‰
8. At a certain region, the electric potential is found to be 𝑽 𝒙, π’š, 𝒛 = πŸ‘π’™πŸ π’š + 𝒛. Calculate the
electric field at point (1,0,1).
Solution
The electric field relates to potential through
𝐸π‘₯ = −
πœ•π‘‰
= −6π‘₯𝑦
πœ•π‘₯
𝐸𝑦 = −
πœ•π‘‰
= −3π‘₯ 2
πœ•π‘¦
𝐸𝑧 = −
Therefore at the point (1,0,1) we get
𝐸π‘₯ = 0
𝐸𝑦 = −3
Hence the electric field is
𝐸 = −3𝑗 − π‘˜
𝐸𝑧 = −1
πœ•π‘‰
= −1
πœ•π‘§
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