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electric field and its applications
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Why does a positive test charge move away from a positive charge?
Why does a positive test charge move towards a negative charge?
What exists around electric charges?
Around electric charges there is an electric field
It is a region surrounding the charge where electric force effects appear
The electric field is a vector quantity with magnitude and direction
Useful information: Calculating electric field magnitude and direction
We place a small positive test charge
at the point where we want to calculate the field
We calculate the electric force acting on the test charge:
\[F_e=\frac{K.Q.q}{r^2}\]
The electric field is the electric force acting on the test charge divided by the test charge at that point:
\[E=\frac{F_e}{q}= \frac{\frac{K.Q.q}{r^2}}{q}=\frac{K.Q}{r^2} \]
It is measured in Newtons/Coulomb
There are two methods to determine direction:
First method: Place a test charge at the point where you want to determine direction and according to vector rules, the product of a number and a vector. If a positive charge is placed, the field direction is the same as the force direction. If the test charge is negative, the field direction is opposite to the electric force direction.
Second method: By drawing field lines for a positive charge and a negative charge
Field lines for a positive charge emerge from the charge
Field lines for a negative charge enter the charge
Electric Field
In this simulation, we place a test charge within the field region and determine the direction of the force acting on the test charge. From the force and test charge, we determine the electric field direction.
1
One of the following units is equivalent to the unit of electric field measurement
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Choose the correct answer
2
An electric field with intensity
\[E=3×10^2\;\;N/C\] has an electron placed inside it.
The electric force acting on the electron equals
\[q_e=1.6 ×10^{-19}C\]
Click here to show solution method
Choose the correct answer
Electric Field Line Mapping
Properties of Electric Field Lines
They are imaginary lines emerging from positive charges and entering negative charges
The density of field lines is directly proportional to the magnitude of the electric charge
The field direction at any point is tangent to the field line at that point
Electric field lines do not intersect
There are two types of electric fields
Uniform field - constant in magnitude and direction at all points, with straight parallel lines
Non-uniform field - varying in magnitude or direction or both
Applications of Electric Fields
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What happens to the potential energy of a ball when we lift it from the ground?
What happens to the electric potential energy of a positive charge when it is moved in the field direction?
What happens to the electric potential energy of a negative charge when it is moved opposite to the field direction?
What happens to the electric potential energy of a negative charge when it is moved in the field direction?
What happens to the electric potential energy of a positive charge when it is moved opposite to the field direction?
The ball loses gravitational potential energy as it approaches the ground and cannot return to its original position
When we move a positive charge in the field direction, it loses electric potential energy and cannot return to its original position
When we move a positive charge opposite to the field direction, it gains electric potential energy and can return to its original position without external force
When we move a negative charge in the field direction, it gains electric potential energy and can return to its original position
When we move a negative charge opposite to the field direction, it loses electric potential energy and cannot return to its original position without external force
Complete the table data below
Change in electric potential energy
Direction of charge movement
Charge type
\[...........\]
In field direction
Positive
\[...........\]
Opposite to field
Positive
\[...........\]
In field direction
Negative
\[...........\]
Opposite to field
Negative
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It is the potential at the final point minus the potential at the initial point
When asked to calculate the potential difference between points
\[A,B\]
It is calculated as follows:
\[ ∆ 𝑉_{𝐴𝐵}=𝑉_B - 𝑉_A\]
The potential at a point is the potential energy stored per unit charge at that point. When asked to calculate the potential at a point in an electric field, we place a test charge at that point, determine the electric potential energy stored by the test charge at that point, and divide by the test charge:
\[∆ 𝑉_{𝐴𝐵}=\frac{PE_B}{q}-\frac{PE_A}{q}=\frac{∆PE_{AB}}{q}\]
\[∆PE_{AB}= W_{A→B} \] Here, the work done in moving the charge from the starting point to the endpoint by an external force
If an internal force is used to do the work, we add a negative sign to the relationship to maintain equality:
\[ ∆ 𝑉_{𝐴𝐵}= \frac{ W_{A→B}}{q}\]
When is the potential difference between two points positive, negative, or zero?
In the figure below, when is the potential difference
\[∆𝑉_{AB}\]
positive, negative, or zero?
To answer, we place a positive or negative test charge
at the starting point and move it to the endpoint
by an external force
which is opposite to the internal force. If the external force is in the direction of displacement, then
the work done is positive (greater than zero).
If the external force is opposite to the displacement, then
the work done is negative (less than zero). If the force is perpendicular to the displacement, then the work is zero. Then we apply the relationship
taking into account the type of test charge:
\[ ∆ 𝑉_{𝐴𝐵}= \frac{ W_{A→B}}{q}\] The potential difference is positive
when \[ ∆ 𝑉_{𝐴𝐵}>0\] The potential difference is negative
when \[ ∆ 𝑉_{𝐴𝐵}<0\]
The potential difference is zero
when \[ ∆ 𝑉_{𝐴𝐵}=0\]
Experiment to measure electric potential of a point charge
Choose the type of field-generating charge and move the potential meter to the desired location
Change the location and observe the measurement again to determine the difference
Measure the potential for points that are equidistant from the charge
What do you conclude?
..................................................................
Measure the potential for several points in the field direction. What do you conclude?
..................................................................
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Measuring Potential Difference Between Two Points in a Uniform Field
Which of the following shapes represents a uniform field and which represents a non-uniform field?
Properties of a uniform field: constant in magnitude and direction at all points, with straight parallel lines equally spaced apart.
Experiment to measure potential difference between capacitor plates (uniform field)
The potential difference equals
\[∆𝑉=1.5 \;\;V \Rightarrow\;\;V^+=1.5 \;\;V\;\;\;,\;\;\;V^-=0V\]
When measuring potential difference between the negative capacitor plate and point
\[A\]\[∆𝑉=1 \;\;V \Rightarrow\;\;V_A^+=1 \;\;V\;\;\;,\;\;\;V^-=0V\] When measuring potential difference between the negative capacitor plate and point
\[B\]\[∆𝑉=0.5 \;\;V \Rightarrow\;\;V_B^+=0.5 \;\;V\;\;\;,\;\;\;V^-=0V\] When measuring potential difference between the negative capacitor plate and point
\[C\]\[∆𝑉=0.5 \;\;V \Rightarrow\;\;V_C^+=0.5 \;\;V\;\;\;,\;\;\;V^-=0V\]
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The potential difference in a uniform electric field is given by: \[ ∆V_{AB}= E . d \]
\[E\] Electric field strength \[d=0.3m \] Displacement parallel to field lines
3
Three points inside a uniform field as shown below. One of the lines represents the correct shape of uniform field lines given that \[ V_A=8V , V_B=6V , V_C=8V \]
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Choose the correct answer
4
In the figure below, a uniform electric field with intensity
\[E=1000\frac{N}{C}\] The potential difference between points \[∆𝑉_{𝑎 𝑑} \]
equals
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Choose the correct answer
Millikan Oil Drop Experiment
This simulation is a simplified version of an experiment conducted by Robert Millikan, where slightly ionized oil droplets were sprayed into an electric field and observed. The voltage was changed until the droplet became balanced and the electric force equaled the weight of the droplet.
Answer the following questions through the experiment:
\[ ..................... \] Click here to show solution method
An oil droplet charged positively - what is the reason for its movement towards the ground?
\[ ...................... \] Click here to show solution method
When the droplet enters the charged box, determine the electric field direction and the forces acting on it.
\[ ..................... \] Click here to show solution method
What is the relationship used to calculate the field intensity inside the box?
\[ ...................... \] Click here to show solution method
Millikan changed the field intensity until the droplet became balanced. Derive the relationship used to calculate the droplet's charge.
Dialogue between physics teacher and student
Student
Physics Teacher
On the outer surface of the conductor
When charging a conductor, where are the charges located?
They don't move because there's no field on the surface. The electric field is perpendicular to the surface.
Do charges move on the conductor's surface and why?
No charges exist due to repulsive forces between charges causing them to separate, with the furthest place being the outer surface.
If the conductor was hollow, would there be charges on the inner surface and why?
Charges accumulate on the pointed tip because they cannot separate
If the conductor had a pointed tip, would charge distribution be uniform and why?
Capacitor Capacitance
Capacitor capacitance: Its ability to hold charge
A capacitor may be uncharged yet we say it has capacitance
This simulation
demonstrates how to charge a capacitor using a battery and the relationship between the potential difference across the capacitor and its charge
Complete the following table based on the experiment
Capacitor Capacity
Calculating Capacitor Capacity
Charge (nC)
Potential Difference (V)
Capacity (nF)
Capacity through the slope of the graph between voltage and charge
Calculate the slope of the graph. What do you conclude from the previous experiment?
\[.......................................................\]
\[.......................................................\] \[.......................................................\]
electric field and its applications |
Click here to show learning outcomes
|
Why does a positive test charge move away from a positive charge?
|
|
Around electric charges there is an electric field
|
We place a small positive test charge
at the point where we want to calculate the field
We calculate the electric force acting on the test charge:
\[F_e=\frac{K.Q.q}{r^2}\]
The electric field is the electric force acting on the test charge divided by the test charge at that point:
\[E=\frac{F_e}{q}= \frac{\frac{K.Q.q}{r^2}}{q}=\frac{K.Q}{r^2} \]
It is measured in Newtons/Coulomb
There are two methods to determine direction:
First method: Place a test charge at the point where you want to determine direction and according to vector rules, the product of a number and a vector. If a positive charge is placed, the field direction is the same as the force direction. If the test charge is negative, the field direction is opposite to the electric force direction.
Second method: By drawing field lines for a positive charge and a negative charge
One of the following units is equivalent to the unit of electric field measurement Choose the correct answer An electric field with intensity
\[E=3×10^2\;\;N/C\] has an electron placed inside it.
The electric force acting on the electron equals
\[q_e=1.6 ×10^{-19}C\]
Choose the correct answer They are imaginary lines emerging from positive charges and entering negative charges
What happens to the potential energy of a ball when we lift it from the ground?
The ball loses gravitational potential energy as it approaches the ground and cannot return to its original position
Change in electric potential energy
Direction of charge movement
Charge type
\[...........\]
In field direction
Positive
\[...........\]
Opposite to field
Positive
\[...........\]
In field direction
Negative
\[...........\]
Opposite to field
Negative
It is the potential at the final point minus the potential at the initial point
When asked to calculate the potential difference between points
\[A,B\]
It is calculated as follows:
\[ ∆ 𝑉_{𝐴𝐵}=𝑉_B - 𝑉_A\]
The potential at a point is the potential energy stored per unit charge at that point. When asked to calculate the potential at a point in an electric field, we place a test charge at that point, determine the electric potential energy stored by the test charge at that point, and divide by the test charge:
\[∆ 𝑉_{𝐴𝐵}=\frac{PE_B}{q}-\frac{PE_A}{q}=\frac{∆PE_{AB}}{q}\]
\[∆PE_{AB}= W_{A→B} \] Here, the work done in moving the charge from the starting point to the endpoint by an external force
If an internal force is used to do the work, we add a negative sign to the relationship to maintain equality:
\[ ∆ 𝑉_{𝐴𝐵}= \frac{ W_{A→B}}{q}\]
In the figure below, when is the potential difference
\[∆𝑉_{AB}\]
positive, negative, or zero?
To answer, we place a positive or negative test charge
at the starting point and move it to the endpoint
by an external force
which is opposite to the internal force. If the external force is in the direction of displacement, then
the work done is positive (greater than zero).
If the external force is opposite to the displacement, then
the work done is negative (less than zero). If the force is perpendicular to the displacement, then the work is zero. Then we apply the relationship
taking into account the type of test charge:
\[ ∆ 𝑉_{𝐴𝐵}= \frac{ W_{A→B}}{q}\] The potential difference is positive
when \[ ∆ 𝑉_{𝐴𝐵}>0\] The potential difference is negative
when \[ ∆ 𝑉_{𝐴𝐵}<0\]
The potential difference is zero
when \[ ∆ 𝑉_{𝐴𝐵}=0\]
Which of the following shapes represents a uniform field and which represents a non-uniform field?
Properties of a uniform field: constant in magnitude and direction at all points, with straight parallel lines equally spaced apart.
Three points inside a uniform field as shown below. One of the lines represents the correct shape of uniform field lines given that \[ V_A=8V , V_B=6V , V_C=8V \] Choose the correct answer In the figure below, a uniform electric field with intensity
\[E=1000\frac{N}{C}\] The potential difference between points \[∆𝑉_{𝑎 𝑑} \] Choose the correct answer
Millikan Oil Drop Experiment
This simulation is a simplified version of an experiment conducted by Robert Millikan, where slightly ionized oil droplets were sprayed into an electric field and observed. The voltage was changed until the droplet became balanced and the electric force equaled the weight of the droplet.
\[ ..................... \] An oil droplet charged positively - what is the reason for its movement towards the ground? \[ ...................... \] When the droplet enters the charged box, determine the electric field direction and the forces acting on it. \[ ..................... \] What is the relationship used to calculate the field intensity inside the box? \[ ...................... \] Millikan changed the field intensity until the droplet became balanced. Derive the relationship used to calculate the droplet's charge. Dialogue between physics teacher and student Student Physics Teacher
Capacitor capacitance: Its ability to hold charge
A capacitor may be uncharged yet we say it has capacitance
This simulation
Calculate the slope of the graph. What do you conclude from the previous experiment?
\[.......................................................\]
\[.......................................................\] \[.......................................................\]
Field lines for a positive charge emerge from the charge
Field lines for a negative charge enter the charge
Electric Field
In this simulation, we place a test charge within the field region and determine the direction of the force acting on the test charge. From the force and test charge, we determine the electric field direction.
Click here to show solution method
Click here to show solution method
The density of field lines is directly proportional to the magnitude of the electric charge
The field direction at any point is tangent to the field line at that point
Electric field lines do not intersect
There are two types of electric fields
Uniform field - constant in magnitude and direction at all points, with straight parallel lines
Non-uniform field - varying in magnitude or direction or both
Applications of Electric Fields
Click here to show learning outcomes
What happens to the electric potential energy of a positive charge when it is moved in the field direction?
What happens to the electric potential energy of a negative charge when it is moved opposite to the field direction?
What happens to the electric potential energy of a negative charge when it is moved in the field direction?
What happens to the electric potential energy of a positive charge when it is moved opposite to the field direction?
When we move a positive charge in the field direction, it loses electric potential energy and cannot return to its original position
When we move a positive charge opposite to the field direction, it gains electric potential energy and can return to its original position without external force
When we move a negative charge in the field direction, it gains electric potential energy and can return to its original position
When we move a negative charge opposite to the field direction, it loses electric potential energy and cannot return to its original position without external force
Click here to show solution method
Experiment to measure electric potential of a point charge
Choose the type of field-generating charge and move the potential meter to the desired location
Change the location and observe the measurement again to determine the difference
Measure the potential for points that are equidistant from the charge
What do you conclude?
..................................................................
Measure the potential for several points in the field direction. What do you conclude?
..................................................................
Click here to show required results
Measuring Potential Difference Between Two Points in a Uniform Field
Experiment to measure potential difference between capacitor plates (uniform field)
The potential difference equals
\[∆𝑉=1.5 \;\;V \Rightarrow\;\;V^+=1.5 \;\;V\;\;\;,\;\;\;V^-=0V\]
When measuring potential difference between the negative capacitor plate and point
\[A\]\[∆𝑉=1 \;\;V \Rightarrow\;\;V_A^+=1 \;\;V\;\;\;,\;\;\;V^-=0V\] When measuring potential difference between the negative capacitor plate and point
\[B\]\[∆𝑉=0.5 \;\;V \Rightarrow\;\;V_B^+=0.5 \;\;V\;\;\;,\;\;\;V^-=0V\] When measuring potential difference between the negative capacitor plate and point
\[C\]\[∆𝑉=0.5 \;\;V \Rightarrow\;\;V_C^+=0.5 \;\;V\;\;\;,\;\;\;V^-=0V\]
Click here to show required results
The potential difference in a uniform electric field is given by: \[ ∆V_{AB}= E . d \]
\[E\] Electric field strength \[d=0.3m \] Displacement parallel to field lines
Click here to show solution method
equals
Click here to show solution method
Answer the following questions through the experiment:
Click here to show solution method
Click here to show solution method
Click here to show solution method
Click here to show solution method
When charging a conductor, where are the charges located?
Do charges move on the conductor's surface and why?
If the conductor was hollow, would there be charges on the inner surface and why?
If the conductor had a pointed tip, would charge distribution be uniform and why?
Capacitor Capacitance
demonstrates how to charge a capacitor using a battery and the relationship between the potential difference across the capacitor and its charge
Complete the following table based on the experiment
Calculating Capacitor Capacity
Charge (nC)
Potential Difference (V)
Capacity (nF)
Capacity through the slope of the graph between voltage and charge
Click here to show the solution method
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