Coulomb's Law Problem: Calculating Force and Direction [Homework Solution]

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The discussion revolves around solving a Coulomb's Law problem involving the calculation of force and direction between point charges. The initial calculations yield a net force of 0.67N, but participants debate the accuracy of this value and the correct method for determining the force vector and its magnitude. After several iterations, a revised force vector calculation suggests a magnitude of 3.84N, with a direction of approximately 10 degrees relative to the x-axis. Additionally, there are inquiries about the effects of dielectric materials on force calculations between charges, leading to discussions about equivalent permittivity and the complexity of the setup. The thread concludes with a request for assistance on a separate problem involving charge distribution between parallel plates.
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Homework Statement


HW1_1.jpg



Homework Equations


F = | kqQ / r*r |


The Attempt at a Solution


Magnitude:
F_net = F_1on3 + F_2on3
F_1on3 = (9E9)(70E-6)(15E-6) / ( (sqrt(5))^2 ) = 1.89N
F_2on3 = (9E9)(-36E-6)(15E-6) / ( (2)^2 ) = -1.215N ~ -1.22N
F_net = 1.89 - 1.22 = 0.67N

Direction = arctan(F_y/F_x)
In component form the vectors for r_13 and r_23 are:
r_13 = 2i + 1j && r_23 = 2j

Therefore the force in component notation would be:
Fvec_13 = 1.89(2i + 1j) = 3.78i + 1.89j
Fvec_23 = -1.22(2j) = -2.44j
Fvec_net = 3.78i - 0.55j

Thus angle must be:
theta = arctan(-0.55/3.78) = -8.13 degrees

I entered 0.675 for the magnitude and was told it was incorrect, although I'm positive I did the work correctly. Am I wrong? (I am a little unsure about the angle part)

Thanks in advance.
 
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One mistake I see is that, for example, r_13 must be converted to a unit vector before you can use it to write Fvec_13. r_23 as a unit vector is just j.
 
So r_13 as a unit vector would be squareroot(5), and r_23 would be squareroot(2), no?
 
I'm not sure what you mean.

r_13 as a unit vector: (2i + 1j) / sqrt(5)

r_23 as a unit vector = j
 
godmoktail said:
Magnitude:
F_net = F_1on3 + F_2on3
F_1on3 = (9E9)(70E-6)(15E-6) / ( (sqrt(5))^2 ) = 1.89N
F_2on3 = (9E9)(-36E-6)(15E-6) / ( (2)^2 ) = -1.215N ~ -1.22N
F_net = 1.89 - 1.22 = 0.67N

Direction = arctan(F_y/F_x)
In component form the vectors for r_13 and r_23 are:
r_13 = 2i + 1j && r_23 = 2j

Therefore the force in component notation would be:
Fvec_13 = 1.89(2i + 1j) = 3.78i + 1.89j
Fvec_23 = -1.22(2j) = -2.44j
Fvec_net = 3.78i - 0.55j

Thus angle must be:
theta = arctan(-0.55/3.78) = -8.13 degrees

I entered 0.675 for the magnitude and was told it was incorrect, although I'm positive I did the work correctly. Am I wrong? (I am a little unsure about the angle part)

Are you sure that's the magnitude?
 
Are you saying that I should have entered 0.67N? Either way, I would like to ensure that my magnitude calculations are correct, regardless of what I entered on my homework software.

If they asked for the Force vector, then I would have to multiply the magnitude by r^ (r hat), no?
 
godmoktail said:
Are you saying that I should have entered 0.67N? Either way, I would like to ensure that my magnitude calculations are correct, regardless of what I entered on my homework software.

If they asked for the Force vector, then I would have to multiply the magnitude by r^ (r hat), no?

What I'm asking is given your statement of the force vector is that it's magnitude?
 
I take Fvec_net as the Force vector 3.78i - 0.55j (with magnitude 0.67N) which I wrote in component notation
 
godmoktail said:
I take Fvec_net as the Force vector 3.78i - 0.55j (with magnitude 0.67N) which I wrote in component notation
F_net = 1.89 - 1.22 = 0.67N

The problem is that this is not the net magnitude. To properly calculate the magnitude of the result you need to have done a vector addition as these magnitudes are not acting along the same line.
 
  • #10
Ok, here is my retry.

r_vec31 = -2i - j
r_vec32 = -2j

r_hat31 = (-2i - j) / sqrt(5)
r_hat32 = (-2j) / 2 = -j

Fvec_31 = 4.23 * r_hat31 N = -3.78i - 1.89j
Fvec_32 = -1.22 * r_hat32 N = 1.22j
Fvec_net = (-3.78i - 0.67j)
Fvec_mag = 3.84 N
So the magnitude would be 3.84N, NO?
 
  • #11
And since the components are -3.78 and 0.67, then the direction is:
arctan(-.67/-3.78) = 10 deg, no?

thanks in advance
 
  • #12
godmoktail said:
Ok, here is my retry.

r_vec31 = -2i - j
r_vec32 = -2j

r_hat31 = (-2i - j) / sqrt(5)
r_hat32 = (-2j) / 2 = -j

Fvec_31 = 4.23 * r_hat31 N = -3.78i - 1.89j
Fvec_32 = -1.22 * r_hat32 N = 1.22j
Fvec_net = (-3.78i - 0.67j)
Fvec_mag = 3.84 N

So the magnitude would be 3.84N, NO?

Yes, that looks good.
 
  • #13
godmoktail said:
And since the components are -3.78 and 0.67, then the direction is:
arctan(-.67/-3.78) = 10 deg, no?

thanks in advance

10 degrees with respect to what? x-axis?
 
  • #14
LowlyPion said:
10 degrees with respect to what? x-axis?

yes, the x-axis
 
  • #15
godmoktail said:
yes, the x-axis

OK. Looks good then.
 
  • #16
hello sir ...can u help me i have a problem...two point charges placed at d distance .two mediums of dielectric constants k1 and k2 are filled between them.one upto distance d1 and other upto d2 distance.what is the force acting between the charge.d1+d2=d.
 
  • #17
I think I would use the equivalent εeq in place of εo

ε1 = k1*εo
ε2 = k2*εo

which I think intuitively can be determined similarly to capacitors in series by

d/εeq = d1/(k1*εo) + d2/(k2*εo)

You may want to satisfy yourself that this is correct.
 
  • #18
sir thanks for replying. sir i have another question also if these mediums are placed one above the other between two point charges .the lengh of mediums is same as the distance between charges .what will be force between the charges .
_____________________________________
q1|________________k2_____________________| q2
|________________k1_____________________|
 
  • #19
r kumar said:
sir thanks for replying. sir i have another question also if these mediums are placed one above the other between two point charges .the lengh of mediums is same as the distance between charges .what will be force between the charges .
_____________________________________
q1|________________k2_____________________| q2
|________________k1_____________________|

That seems to be more complicated because the thicknesses are unknown. If the thicknesses are large and the boundary bisects the point charges then by symmetry I would suppose that

εeq = 1/2*(k1 + k2)εo

BUT for thicknesses less than d between the charges, ... that looks more complicated than I want to consider.
 
  • #20
sir both the mediums have thickness same as that of d distance between the chages.
 
  • #21
r kumar said:
sir both the mediums have thickness same as that of d distance between the chages.

Then ... more or less the average is what I'd say.
 
  • #22
thanks sir for helping me
 
  • #23
sir i have a prblm can u help me :there are three plates placed parallel to each other distance between 1nd @2nd is D distance between 2nd @ 3rd is 2D .1st plate given Q charge 2nd plate given 2Q.1st @ 2nd are connected by wire through a key .iniatlly key was open .if key is closed what amount of charge will pass through key .sir pls reply i will be very thankfull to u.
 

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