Calculate Net Electric Field at Point P from Two Charges 45 cm Apart

[x86]

Homework Statement

Two charges are 45 cm apart. The charge on q1 is 3.3 * 10^-9 C and the charge on q2 is -1 * 10^-8 C.

a) Calculate the net electric field at point p, 27 cm from the positive charge.

Homework Equations

Ef = kq1 / r^2
k = 9.0 * 10^9 N m^2 C^-2

The Attempt at a Solution

Okay, so there are two force vectors

+...p...-

p is 27 cm to the right of the + charge
p is 45-27 = 18 cm to the left of the - charge

Ef1 = 9*10^9*3.3*10^-9 / (27/100)^2 = 407.4 N/C
Ef2 = 9*10^9*-1*10^-8 / (18/100)^2 =-2777.8 N/C

So we have two vectors
--------> (407.4 N/C)
<------- (-2777.8 N/C)

If we add them, we get an answer around 2300, but this is wrong. The answer is 2777+407 but the answer that I get is 407 - 2777.8

Now my question is, how come the book adds the forces together? I am confused

 

[TSny]

So we have two vectors
--------> (407.4 N/C)
<------- (-2777.8 N/C)

Did you get the directions of both vectors correct? Remember, q2 is a negative charge.

 

[Curious3141]

Homework Statement

Two charges are 45 cm apart. The charge on q1 is 3.3 * 10^-9 C and the charge on q2 is -1 * 10^-8 C.

a) Calculate the net electric field at point p, 27 cm from the positive charge.

Homework Equations

Ef = kq1 / r^2
k = 9.0 * 10^9 N m^2 C^-2

The Attempt at a Solution

Okay, so there are two force vectors

+...p...-

p is 27 cm to the right of the + charge
p is 45-27 = 18 cm to the left of the - charge

Ef1 = 9*10^9*3.3*10^-9 / (27/100)^2 = 407.4 N/C
Ef2 = 9*10^9*-1*10^-8 / (18/100)^2 =-2777.8 N/C

So we have two vectors
--------> (407.4 N/C)
<------- (-2777.8 N/C)

If we add them, we get an answer around 2300, but this is wrong. The answer is 2777+407 but the answer that I get is 407 - 2777.8

Now my question is, how come the book adds the forces together? I am confused

Remember that the electric field at a point is the electrostatic force acting on a unit *positive* charge at that point. Since force is a vector quantity, it has both magnitude and direction.

The positive charge q1 repels the positive unit charge, whereas the negative charge q2 attracts it. Therefore, both forces are acting in the same direction (away from q1, towards q2). That's why you should add the magnitudes.

 

[x86]

Remember that the electric field at a point is the electrostatic force acting on a unit *positive* charge at that point. Since force is a vector quantity, it has both magnitude and direction.

The positive charge q1 repels the positive unit charge, whereas the negative charge q2 attracts it. Therefore, both forces are acting in the same direction (away from q1, towards q2). That's why you should add the magnitudes.

Thank you. However, there is no positive charge at point p. So as far as I know nothing is being attracted or repelled over there?

 

[ehild]

Draw the direction of the electric fields at P due to both charges. What is the direction of the electric field around a positive charge and what is it around a negative charge? Remember, how the electric field lines were drawn in your notes or book.

ehild

 

[x86]

Okay, thank you. I think I am understanding it now.

At point P, the positive charge is being attracted to the negative charge so we have the vector
----------> 407N/C

Also, the negative charge is is being attracted to by the positive charge again, so we have the force
---------->2777N/C

Is that reasoning correct?

Also, are we assuming point P is positively charged? Because if we are, that would change my reasoning.

Thanks

 

[ehild]

By definition, the electric field at a point is equal to the force acting on a positive unit charge placed at that point.

If you place a positive unit charge at P, the positive charge (the red one) repels it, the force exerted on P from the positive charge points to the right.

The negative charge (blue) attracts the charge at P, so its force points to the right again.


ehild