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Electric Potential Energy	Gravitational Potential Energy
What happens when we place a positive or negative charge in the field of another charge	What happens when we place a ball far from Earth's surface in Earth's field
The charge gains an amount of energy that depends on its position relative to surrounding charges, called electric potential energy	The object gains an amount of energy due to its position relative to Earth, called gravitational potential energy
It becomes capable of moving on its own because it possesses energy	It becomes capable of moving on its own because it possesses energy
\[Ue=K\frac{Q.q}{r}\]	\[UG=m.g.h\]
The electric potential energy of a charge increases when moved and becomes capable of returning to its original position. If it cannot return, we say the potential energy has decreased	The gravitational potential energy of an object increases when moved away from the surface and becomes capable of returning to its original position. If it cannot return, we say the potential energy has decreased

Complete the following table data

(Remains constant - Increases - Decreases) Change in electric potential energy	(Perpendicular to field - In field direction - Opposite to field direction) Direction of charge movement	(Negative - Positive) Charge type
Potential energy decreases	In field direction	\[..................\]
\[..............\]	Opposite to field	Positive
Potential energy increases	\[................\]	Negative
\[..............\]	Opposite to field	Negative
\[..............\]	Perpendicular to field	Negative - Positive

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Previous information
\[1)E= K+U\] Total energy = Kinetic energy + Potential energy \[2) ∆𝐾+∆𝑈=0 , ∆𝐾=−∆𝑈\] In a closed system, total energy remains constant throughout motion
\[3) W= ∆𝐾\] Work done in moving an object equals change in kinetic energy \[4) W=- ∆U\] Work done in moving an object equals negative change in potential energy \[ 5) W=F.d.Cos (𝜃 )\] Work is positive if force or one of its components is in displacement direction
Work is negative if force or one of its components is opposite to displacement
Work is zero if force is perpendicular to displacement

Change in electric potential energy in a uniform field

(Electric Potential Energy)

1. Definition of Electric Potential Energy

Electric potential energy is the energy stored in a system of charges due to their relative positions in an electric field.

2. Change in Potential Energy in Uniform Field

In a uniform electric field, the change in potential energy when moving a charge a distance is calculated by:
\[ ∆U=-W=-F.d.Cos (𝜃 )=- q.E.d.Cos (𝜃 )\]

3. Derivation of the Law

The law is derived from the concept of work done:

1. Work (W) needed to move the charge: W = F × d
2. Electric force: F = qE
3. Substituting: W = qEd
4. According to energy conservation: ΔU = -W
5. Thus: ΔU = qEd (considering sign)

4. Purpose of Study

• Understand energy storage mechanism in electric fields
• Analyze energy transfer in electric circuits
• Design energy storage systems like capacitors

5. Practical Applications

Application	Description
Capacitors	Energy storage between plates
Cathode Ray Tube	Electron acceleration in CRT screens
Batteries	Convert chemical energy to electrical energy

\[ ∆U=-W=-F.d.Cos (𝜃 )=- q.E.d.Cos (𝜃 )\]

\[∆U= - q.E.d\] \[∆U\] Change in potential energy in uniform field, a scalar quantity measured in Joules \[q\] Magnitude and type of moved charge \[E\] Field magnitude \[d\] Displacement parallel to field lines If in field direction takes positive sign If opposite to field takes negative sign

1

A positive charge (proton) is moved between two points \[A\;\;\Rightarrow\;\;B\] Coordinates of the points \[A=(2\;m,-3\;m)\;\;\;\;\;\;\;B=(-2\;m,2\;m)\] In a uniform field of intensity \[E=4× 10^3\;\;N/C\]
Directed toward the negative vertical axis as shown below. The change in electric potential energy equals \[ q_p = 1.6 ×10^{-19}C\]

∆ 𝑈= - 3.2 × 10-15j-C	∆ 𝑈= - 2.56 × 10-15j -A
∆ 𝑈= 2.56 × 10-15j -D	∆ 𝑈= 3.2 × 10-15j -B

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Dipole in a Uniform Field
Self-reading
The change in potential energy equals the work done by internal forces > \[∆U= - q.E.d\] If the dipole charge were zero, there would be two equal charges in magnitude and opposite in type
So the change in potential energy would be zero, meaning it couldn't store energy, but this is incorrect
The dipole has torque \[\vec τ = \vec P .\vec E \] \[W=\int{\vec τ(ɵ).d\vec ɵ}=\int_{{\,ɵ_0}}^{{\,ɵ}}-p. E .Sinɵ .dɵ =-p.E\int_{{\,ɵ_0}}^{{\,ɵ}}Sinɵ.dɵ=P.E(Cosɵ-Cosɵ_0)\] \[U= p.E.Cosɵ=\vec p.\vec E\] The graph below shows the relationship between the dipole's potential energy and the angle between the field and the dipole direction Electric Potential

Electric Potential

Definition: The work needed to move a unit positive charge from infinity to the desired point.

Electric Potential for a Point Charge

\[ V = k \frac{Q}{r}\] ➔ \[ V = (8.99×10^9)×\frac { Q }{ r}\]

Properties of Electric Potential:

• Scalar quantity (has magnitude only)
• Depends on the point's distance from the charge
• Measured in Volt (V)

Practical Example:

A point charge \[ Q = +5\;nC\] Find the electric potential at a distance \[r= 0.3\;m \]

Calculating the field (E):
\[V = (8.99×10^9) × \frac {5×10^{-9}}{0.3}\]
\[V = 149.83 V \]

Point Charge Electric Potential Calculator

Charge amount (µC):

Charge type: Positive Negative

Distance from charge (m):

Calculate Potential

Notes:

• Potential equation: \[V =\frac { k * Q }{ r}\]
• k: Coulomb's constant \[k=(8.99×10^9)N.m^2/c^2\]
• Q: Charge value in Coulombs
• r: Distance from charge in meters
• Result appears in scientific notation (e.g., 1.23e+5)

Electric Potential for Multiple Point Charges

Basic Concept:

Electric Potential: At a point in space is the amount of work needed to move a positive test charge from infinity to that point without acceleration.

or Electric Potential: is the electric potential energy per unit charge

Calculating Potential for a Single Point Charge:

\[ V =k *\frac { q }{ r}\]

Where:
- k is Coulomb's constant
\[k=(8.99×10^9)N.m^2/c^2\] - q is charge magnitude (Coulombs)
- r is distance from charge to studied point (meters)

For Multiple Point Charges:

\[ V_{total} = Σ V_i = k Σ\frac {q_i}{r_i}\]

Total potential is calculated by algebraic sum of individual potentials for each charge (superposition principle)

Calculation Importance:

• Designing electronic circuits and electrical complexes
• Analyzing charge systems in capacitors
• Studying electric fields in biological systems (e.g., ECG devices)
• Applications in X-ray systems and microelectronics

Practical Applications:

1. Designing electrostatic storage devices
2. Calculating capacitor capacitance
3. Analyzing heart electrical signals
4. Lightning protection systems
5. Electron beams in cathode ray tubes

Important Notes:

- Potential is a scalar quantity (not a vector) → easier to calculate
- Positive charges produce positive potentials
- Negative charges produce negative potentials
- Unit of measurement: volt (1 volt = 1 joule/coulomb)

Electric Potential Calculator

Charge (µC): Distance (m):

Add Charge ➕ Calculate Potential ⚡

Solved Example

In the figure below, find the potential at the midpoint between the two charges

The point is subject to two potentials

Potential from first charge \[V_1=K.\frac{q_1}{r_1}=9×10^9\frac{-8×10^{-9}}{0.2}=-360 V\] Potential from second charge \[V_2=K.\frac{q_2}{r_2}=9×10^9\frac{-3×10^{-9}}{0.2}=-135 V\]

\[V_{net}=V_1+V_2=-360+(-135)=-495 V\]

Result: If the point where potential is calculated is in the field of more than one charge, find the algebraic sum of potentials affecting that point \[V=K\sum \frac{q_i}{r_i}\]

2

The potential at point A was calculated
\[A\;\;\;\; V_A= -45 V \] and the field at the same point was calculated to be \[E=112.5 \;\;N/C\] Then the magnitude and type of charge affecting that point equals

\[ q= -2 ×10^{-9}\;C\;\;\;\;\;\;-C\]	\[ q= 4 × 10^{-9}\;C \;\;\;\;\;\;-A\]
\[ -q= 4 × 10{-9}\;C\;\;\;\;\;\;-D\]	\[ q= 2 × 10^{-9}\;C\;\;\;\;\;\;-B\]

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Potential Difference Between Two Points

The potential difference between two points is defined as: the work done to move a unit charge from one point to the other \[∆V=V_f-V_i=\frac{U_f}{q}-\frac{U_i}{q}=\frac{∆U}{q}\] \[∆V=-\frac{ W_{A→B}}{q}\] Solved Example

A proton moved between two points in an electric field starting with velocity at point \[A\]\[v=40 \frac{m}{s}\] (B) and its velocity at point B became \[v=10^4\frac{m}{s}\] Calculate the potential difference between the two points knowing that \[q_p=1.6×10^{-19}C , m_p=1.67×10^{-27}kg\]

Solution

\[∆k=k_f-k_i=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2\] \[∆k=\frac{1}{2}×1.67×10^{-27}×(10^4)^2 -\frac{1}{2}×1.67×10^{-27}×(40)^2 =8.34×10^{-20}\frac{Kg.m^2}{s^2}\]

\[∆k=W\]

\[∆V=-\frac{W}{q}=-\frac{8.34×10^{-20}}{1.6×10^{-19}}=-0.52 V\]

Equipotential Surfaces

It's possible for a charge to move between two points in an electric field without work being done

This happens when we move a charge between two points at the same potential - no work is done on the charge because \[∆V=0\] \[∆V = -\frac{W}{q}=0 \]\[W=0\] Points with the same potential

1

Points at the same distance from a positive or negative point charge have the same potential

2

Points located perpendicular to the lines of a uniform field have the same potential

3

Two point charges of different types - equipotential points resemble those of single charges but the field differs in the region between the charges

4

Two point charges of different types Equipotential points exist around each charge The shape of equipotential points and the field differs in the region between the charges

Equipotential Points Experiment In this simulation, you can adjust the charge and position of two charges using sliders or input boxes.
The sliders work, but not smoothly due to calculation complexity
- so it might be better for you to use input boxes.
Choose 3D view and the electric potential appears as a third dimension. Choose equipotential view and you'll see a 2D display with equipotential lines shown. In this view,
you can also choose to see vectors showing the electric field direction

Electric Potential for Different Charge Distributions

Potential: is the work needed to move a test charge from infinity to the desired point divided by the test charge \[∆V=V_f-V_i=-\frac{W_{∞→r}}{q}=-\int_{{\,∞}}^{{\,r}}\frac{\vec F .ds}{q}=-\int_{{\,∞}}^{{\,r}}\frac{ q.\vec E .ds}{q}\] \[∆V=-\int_{{\,∞}}^{{\,r}}{ \vec E .ds}=-\int_{{\,∞}}^{{\,r}}{ \frac{kq}{r^2} .dr}= -k.q\int_{{\,∞}}^{{\,r}}{ \frac{1}{r^2} .dr}= \frac{k.q}{r}\] \[∆V=V_r-V_∞=V_r-0=\frac{k.q}{r}\] \[V_r=\frac{k.q}{r}\]

3

In the figure below, two charges \[q_1= -4 \;\;nC\;\;\;,\;\;\;q_2=6\;\;nC\] The first negative charge is placed at the origin and the second positive charge is placed at a point 0.4 m from the origin The point of zero potential on the line connecting the two charges is located at position

\[ X=0.14 \;\;m\;\;\;\;\;\;-C\]	\[ X=0.18\;\;m \;\;\;\;\;\;-A\]
\[ X=0.12 \;\;m\;\;\;\;\;\;-D\]	\[ X=0.16 \;\;m\;\;\;\;\;\;-B\]

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Electric Potential Difference in a Uniform Field

\[∆V_{AB}=-\frac{W_{A→B}}{q}=-\frac{q.E.d}{q}\] \[∆V_{AB}=-E.d\]

4

In the figure below, a uniform field with intensity \[ E = 500 \;\;N/C \] Then the potential difference between points \[∆𝑉_{𝐴𝐵}=?\] and the dimensions are shown on the drawing equals

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Choose the correct answer

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∆𝑉(𝐴𝐵)=150 V

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∆𝑉(𝐴𝐵)=250 V

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∆𝑉(𝐴𝐵)=200 V

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∆𝑉(𝐴𝐵)=-150 V

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Choose Answer

Solved Example

( λ= 4×10^-8C/m ) A wire uniformly charged along its length with linear charge density
( R ) bent into a semicircle of radius
Calculate the potential at the center of curvature of the wire
Solution
\[V=\int_{{\,0}}^{{\,𝜋R}}\frac{K.dq}{R}\]\[V=\int_{{\,0}}^{{\,𝜋R}}\frac{K.λ.dL}{R}\] \[V=\frac{K.λ}{R}\int_{{\,0}}^{{\,𝜋R}}dL=\frac{K.λ}{R} |𝐿|_{{\,0}}^{{\,𝜋R}}=\frac{K.λ}{R}.[𝜋R-0]\] \[V=K.λ.𝜋\]\[V=9×10^9×4×10^{-8}×𝜋=1131 V\]

Finding Electric Field from Electric Potential From work: \[dW=q.\vec E.\vec {dS}\] \[-q.dV=q.\vec E.\vec {dS}\] \[E=-\frac{dV}{dS}\] \[E_X=-\frac{dV}{dX}, E_Y=-\frac{dV}{dY} , E_Z=-\frac{dV}{dZ}\]
Solved Example

The electric potential in a region of space is given by \[V(X,Y,Z)=5X^2+8XY+7Z\]
( 5,-2,-3 ) then the field value at a point with coordinates
equals
Solution
\[E_X=-(\frac{dV}{dX})=-(\frac{5X^2+8XY+7Z}{dX})=-(10X+8Y)=-(10×5+8×-2)=-34\] \[E_Y=-\frac{dV}{dY}=-(\frac{5X^2+8XY+7Z}{dY})=-(8X)=-(8×5)=-40\] \[E_Z=-\frac{dV}{dZ}=-(\frac{5X^2+8XY+7Z}{dZ})=-7\] \[E(-34 X,-40Y,-7Z)\] \[E=\sqrt {34^2+40^2+7^2}=52.9 N/C\]

Potential Energy of a System of Point Charges From potential definition: \[V=\frac{U}{q}\]\[U=q.V=q.\frac{kQ}{r}\] \[U=k.\frac{Q.q}{r}\] Note: The electric potential energy of a charge at infinity equals zero We have three charges located at the vertices of a right triangle as shown below
Required to calculate the potential energy of the system consisting of three charges
Initially we assume the charges are separated at infinity
We bring the first charge from infinity - potential energy for it alone \[U=0\] (q1 ) When bringing the second charge from infinity next to \[U=K\frac{q_1.q_2}{r_1}\] ( q1, q2 )When bringing the third charge from infinity next to \[U=K\frac{q_1.q_2}{r_1}+K\frac{q_1.q_3}{r_2}+K\frac{q_2.q_3}{\sqrt {{(r_1)^2}+{(r_2)^2}}}\] Which is the electric potential energy of the system Write a comment, and if there is mistake, write and specify its location Write a comment, and if there is mistake, write and specify its location

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