Calculate Magnitude of Force Exerted on q1 by q3 & Horiz. Comp.

In summary, the magnitude of force exerted on charge q1 by charge q3 and horizontal component of force exerted is calculated by using the formula q1q2 / 4piEor^2. For part 1, the answer should be negative due to the force being attractive. For part 2, the vector forces exerted by all the other charges should be calculated and summed up to find the resultant force. If q1 is replaced by a -4*10^-6 C, then the partial forces should all change signs. The original question asks for the resultant force exerted on q1 by the other 3 charges, using vector addition to find the resultant force or by calculating the magnitudes of each force and
  • #1
Bama
15
0
Ok, here is what I have. See attachment to complete the problem.

What is the magnitude of force exerted on charge q1 by charge q3 and horizontal component of force exerted?

Formula: q1q2 / 4piEor^2

Part 1) (4*10^-6)(2.12*10^-6) / 12.56 (8.85*10^-12)(1*10^-2) = 7.628917917176 N or Rounded off as 7.63N

Part 2) 7.63 N sin 135 degrees = 5.39 N

Now I understand the above, howver what I can understand is the following. The resultant force exerted on the charge q1 by the other 3 charges is what? But what if we replaced q1 with charge -4*10^-6, what resultant force would be exerted by the other 3 charges?
 

Attachments

  • Q1 & q2.JPG
    Q1 & q2.JPG
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  • #2
For part 1, your answer should be negative since the force is attractive. As for part 2, calculate the vector forces exerted by all the other charges and sum them all up to find the resultant force. If q1 is replaced by a -4*10^-6 C, then the partial forces you calculated to find the resultant force should all change signs.
 
  • #3
Corneo if my answer suppose to be negative by force of attractive, than I am confused. The choices that I have doesn't have a negative solution at all. So is my equation set up incorrctly and if so, how should it be set up?
 
  • #4
Your formula is correct but you forgot to inclue the sign of q3 in your calculation.

Recall that the electrical force between two charges is given by:
[tex]F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1Q_2}{r^2}[/tex]

So for part 1) Q1 = 4x10^-6 C and Q2 = -2.12x10^-6 C.

Perhaps the choices for the possible solutions are wrong. Unless the question meant the unsigned magnitude of the force.
 
  • #5
The original question is follow. "What is the magnitude of force exerted on charge q1 by charge q2?

choices: 1) 0 N 3) 3.77 N 5) 7.63 N
2) 0.108 N 4) 5.39 N 6) 10.8 N

The question I am confused about is : "What is the resultant force exerted on charge q1, by the other 3 charges? ( Hint: Note angles carefully)." It also states to selected my answer from one of the above choices.

Also you have Q2 as equaling -2.12x10^-6 C, but as I look at the image only q3 and q4 is eqaul to -2.12x10^-6 C and q2 is equal to -3 x 10^-6. Am I mis-understanding what you meant for your Q2 above?
 
  • #6
Well I used [itex]Q_1 \text{ and } Q_2[/itex] as variables in general for the formula. Applied in this case, [itex]Q_1 = q_1[/itex] and [itex]Q_2 = q_3[/itex]. To answer you last question you can use vector addition to find the resultant force. The formula is

[tex]\vec F = \frac {1}{4 \pi \epsilon_0} \frac {Q_1Q_2}{r^2} {\vec r}[/tex]

However you could also just calculate the magnitudes of each force exerted on q1, then calculate the components of each vector and sum them all up.
 

What is the equation for calculating the magnitude of force exerted on q1 by q3?

The equation for calculating the magnitude of force exerted on q1 by q3 is F = (k * q1 * q3) / d2, where k is the Coulomb's constant, q1 and q3 are the magnitudes of the charges, and d is the distance between the two charges.

How do you calculate the horizontal component of the force?

The horizontal component of the force can be calculated by multiplying the magnitude of the force by the cosine of the angle between the force vector and the horizontal axis.

What unit is used to measure force?

The unit used to measure force is the Newton (N). Other common units of force include pounds (lbs) and dynes (dyn).

Can the magnitude of force exerted on q1 by q3 be negative?

Yes, the magnitude of force exerted on q1 by q3 can be negative if the charges have opposite signs and are attracting each other.

How does the distance between the charges affect the magnitude of force?

The distance between the charges is inversely proportional to the magnitude of force. This means that as the distance increases, the force decreases and vice versa.

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