Using Coulomb's Law - AP Physics C: Electricity and Magnetism
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A point charge of 
exerts a force of 
on another charge with 
. How far apart are the two charges?
A point charge of
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To find the distance between the two charges, use Coulomb's Law.
Since we want to find distance, 
, we solve for 
.
We know the values of the force and the two charges.
We can plug in these values and solve for the distance.
To find the distance between the two charges, use Coulomb's Law.
Since we want to find distance,
We know the values of the force and the two charges.
We can plug in these values and solve for the distance.
We have a point charge of 
. Determine the electric field at a distance of 
away from that charge.
We have a point charge of
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Coulomb's law for the electric field from point charges is 
, where we know the values of the following variables.
Using these values, we can solve for the electric field.
Coulomb's law for the electric field from point charges is
Using these values, we can solve for the electric field.
Two positive point charges of 
and 
are place at a distance 
away from each other, as shown below. If a positive test charge, 
, is placed in between, at what distance away from 
will this test charge experience zero net force?
Two positive point charges of
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To find the location at which the test charge experience zero net force, write the net force equation as 
, where 
is the force on the test charge from 
, and 
is the force on the same test charge from 
. Using Coulomb's law, we can rewrite the force equation and set it equal to zero.
In this equation, the distance, 
, is how far away the test charge is from 
, while 
represents how far away the test charge is from 
. Now, we simplify and solve for 
.
Cross-multiply.
%5E2=kq_2qr%5E2 "kq_$1q(d-r)^2$=kq_$2qr^2$")
We can cancel 
and 
. We do not need to know these values in order to solve the question.
Now that we have isolated 
, we can plug in the values given in the question and solve.
To find the location at which the test charge experience zero net force, write the net force equation as
In this equation, the distance,
Cross-multiply.
We can cancel
Now that we have isolated
You are standing on top of a very large positively charged metal plate with a surface charge of 
.
Assuming that the plate is infinitely large and your mass is 
, how much charge does your body need to have in order for you to float?
You are standing on top of a very large positively charged metal plate with a surface charge of
Assuming that the plate is infinitely large and your mass is
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Consider the forces that are acting on you. There is the downward (negative direction) force of gravity, 
. In order for you to float, there has to be an upward (positive direction) force, and that upward force is coming from the metal plate, 
. To show that you would float, the net force equation is written as 
, where 
is the charge on you.
For plates that are charged, know that 
.
Knowing this, the force equation becomes 
.
Solve for 
.
Now we can plug in our given values, and solve for the charge.
Consider the forces that are acting on you. There is the downward (negative direction) force of gravity,
For plates that are charged, know that
Knowing this, the force equation becomes
Solve for
Now we can plug in our given values, and solve for the charge.
What is the electric force between two charges, 
and 
, located 
apart?
What is the electric force between two charges,
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The equation for finding the electric force between two charges is 
, where 
. Using this, we can rewrite the force equation.
Now, we can use the values given in the question to solve for the electric force between the two particles.
The equation for finding the electric force between two charges is
Now, we can use the values given in the question to solve for the electric force between the two particles.
What is the magnitude of the electric field at a field point 
from a point charge of 
?
What is the magnitude of the electric field at a field point
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The equation to find the strength of an electric field is 
.
We can use the given values to solve for the strength of the field at a distance of 
.
The equation to find the strength of an electric field is
We can use the given values to solve for the strength of the field at a distance of
Two capacitors are in parallel, with capacitance values of 
and 
. What is their equivalent capacitance?
Two capacitors are in parallel, with capacitance values of
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The equivalent capacitance for capacitors in parallel is the sum of the individual capacitance values.
Using the values given in the question, we can find the equivalent capacitance.
The equivalent capacitance for capacitors in parallel is the sum of the individual capacitance values.
Using the values given in the question, we can find the equivalent capacitance.
A proton moves in a straight line for a distance of 
. Along this path, the electric field is uniform with a value of 
. Find the force on the proton.
The charge of a proton is 
.
A proton moves in a straight line for a distance of
The charge of a proton is
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The force of an electric field is given by the equation 
, where 
is the charge of the particle and 
is the electric field strength. We can use the given values from the question to solve for the force.
The force of an electric field is given by the equation
Two point charges, 
and 
are separated by a distance 
.
The values of the charges are:
The distance is 4.0cm. The point 
lies 1.5cm away from 
on a line connecting the centers of the two charges.
What is the magnitude and direction of the net electric field at point 
due to the two charges?
Two point charges,
The values of the charges are:
The distance is 4.0cm. The point
What is the magnitude and direction of the net electric field at point
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At point 
, the electric field due to 
points toward 
with a magnitude given by:
At point P, the electric field due to Q2 points away from Q2 with a magnitude given by
The addition of these two vectors, both pointing in the same direction, results in a net electric field vector of magnitude 152000 volts per meter, pointing toward 
.
At point
At point P, the electric field due to Q2 points away from Q2 with a magnitude given by
The addition of these two vectors, both pointing in the same direction, results in a net electric field vector of magnitude 152000 volts per meter, pointing toward
A point charge of exerts a force of on another charge with . How far apart are the two charges?
A point charge of
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To find the distance between the two charges, use Coulomb's Law.
Since we want to find distance, , we solve for .
We know the values of the force and the two charges.
We can plug in these values and solve for the distance.
To find the distance between the two charges, use Coulomb's Law.
Since we want to find distance,
We know the values of the force and the two charges.
We can plug in these values and solve for the distance.
We have a point charge of . Determine the electric field at a distance of away from that charge.
We have a point charge of
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Coulomb's law for the electric field from point charges is , where we know the values of the following variables.
Using these values, we can solve for the electric field.
Coulomb's law for the electric field from point charges is
Using these values, we can solve for the electric field.
Two positive point charges of and are place at a distance away from each other, as shown below. If a positive test charge, , is placed in between, at what distance away from will this test charge experience zero net force?
Two positive point charges of
Tap to see back →
To find the location at which the test charge experience zero net force, write the net force equation as , where is the force on the test charge from , and is the force on the same test charge from . Using Coulomb's law, we can rewrite the force equation and set it equal to zero.
In this equation, the distance, , is how far away the test charge is from , while represents how far away the test charge is from . Now, we simplify and solve for .
Cross-multiply.
We can cancel and . We do not need to know these values in order to solve the question.
Now that we have isolated , we can plug in the values given in the question and solve.
To find the location at which the test charge experience zero net force, write the net force equation as
In this equation, the distance,
Cross-multiply.
We can cancel
Now that we have isolated
You are standing on top of a very large positively charged metal plate with a surface charge of .
Assuming that the plate is infinitely large and your mass is , how much charge does your body need to have in order for you to float?
You are standing on top of a very large positively charged metal plate with a surface charge of
Assuming that the plate is infinitely large and your mass is
Tap to see back →
Consider the forces that are acting on you. There is the downward (negative direction) force of gravity, . In order for you to float, there has to be an upward (positive direction) force, and that upward force is coming from the metal plate, . To show that you would float, the net force equation is written as , where is the charge on you.
For plates that are charged, know that .
Knowing this, the force equation becomes .
Solve for .
Now we can plug in our given values, and solve for the charge.
Consider the forces that are acting on you. There is the downward (negative direction) force of gravity,
For plates that are charged, know that
Knowing this, the force equation becomes
Solve for
Now we can plug in our given values, and solve for the charge.
What is the electric force between two charges, and , located apart?
What is the electric force between two charges,
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The equation for finding the electric force between two charges is , where . Using this, we can rewrite the force equation.
Now, we can use the values given in the question to solve for the electric force between the two particles.
The equation for finding the electric force between two charges is
Now, we can use the values given in the question to solve for the electric force between the two particles.
What is the magnitude of the electric field at a field point from a point charge of ?
What is the magnitude of the electric field at a field point
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The equation to find the strength of an electric field is .
We can use the given values to solve for the strength of the field at a distance of .
The equation to find the strength of an electric field is
We can use the given values to solve for the strength of the field at a distance of
Two capacitors are in parallel, with capacitance values of and . What is their equivalent capacitance?
Two capacitors are in parallel, with capacitance values of
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The equivalent capacitance for capacitors in parallel is the sum of the individual capacitance values.
Using the values given in the question, we can find the equivalent capacitance.
The equivalent capacitance for capacitors in parallel is the sum of the individual capacitance values.
Using the values given in the question, we can find the equivalent capacitance.
A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of . Find the force on the proton.
The charge of a proton is .
A proton moves in a straight line for a distance of
The charge of a proton is
Tap to see back →
The force of an electric field is given by the equation , where is the charge of the particle and is the electric field strength. We can use the given values from the question to solve for the force.
The force of an electric field is given by the equation
Two point charges, and are separated by a distance .
The values of the charges are:
The distance is 4.0cm. The point lies 1.5cm away from on a line connecting the centers of the two charges.
What is the magnitude and direction of the net electric field at point due to the two charges?
Two point charges,
The values of the charges are:
The distance is 4.0cm. The point
What is the magnitude and direction of the net electric field at point
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At point , the electric field due to points toward with a magnitude given by:
At point P, the electric field due to Q2 points away from Q2 with a magnitude given by
The addition of these two vectors, both pointing in the same direction, results in a net electric field vector of magnitude 152000 volts per meter, pointing toward .
At point
At point P, the electric field due to Q2 points away from Q2 with a magnitude given by
The addition of these two vectors, both pointing in the same direction, results in a net electric field vector of magnitude 152000 volts per meter, pointing toward
A point charge of exerts a force of on another charge with . How far apart are the two charges?
A point charge of
Tap to see back →
To find the distance between the two charges, use Coulomb's Law.
Since we want to find distance, , we solve for .
We know the values of the force and the two charges.
We can plug in these values and solve for the distance.
To find the distance between the two charges, use Coulomb's Law.
Since we want to find distance,
We know the values of the force and the two charges.
We can plug in these values and solve for the distance.
We have a point charge of . Determine the electric field at a distance of away from that charge.
We have a point charge of
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Coulomb's law for the electric field from point charges is , where we know the values of the following variables.
Using these values, we can solve for the electric field.
Coulomb's law for the electric field from point charges is
Using these values, we can solve for the electric field.