Magnitude of the electric field

In summary: Just remember to switch the symbols as mentioned in the previous comment, and also remember to include the r in the integration. So your final result would be:E = (2kQ) / (Lr)In summary, the magnitude and direction of the electric field at the origin of a semi circle formed by a uniformly charged plastic rod with length L = .24 cm and total charge of 12 x 10^-6 C is (2kQ) / (Lr), with the direction being -j hat. The integration of the arc length and angle is necessary to find the correct value of the electric field.
  • #1
Rijad Hadzic
321
20

Homework Statement


http://imgur.com/a/g5acE

# 34

A plastic rod of length L = .24 cm is uniformly charged with a total charge of 12 x 10^-6 C.

The rod is formed into a semi circle with its center at the origin of the xy plane. What are the magnitude and direction of the electric field at the origin?

Homework Equations


kQ/r^2 = E

The Attempt at a Solution



Ok so can anyone tell me if this statement is correct:

1) Because I'm given L = .24 m, and the circumfrence of a circle is 2pir, since I have a semi circle I can set .24m = pi r, now r = .24m/pi

Going forward assuming that is true...:

The direction of the field is -j hat. Because the x components cancel out.

I consider a small piece of the rod I will call ds that has charge dQ. [itex] dQ = \Lambda ds [/itex] correct?

Anyways the magnitude of the electric field at a, E, is going to be 2dEsin(theta) because its getting its Y component from both sides of the Y axis.so I have E = 2dEsin(theta)

dE = kdQ / r^2

so I have

E = 2*k*dq*sin(thetha) / r^2

but [itex] dQ = \Lambda ds [/itex]

so

[itex] E = (2*k*\Lambda * ds * sin(theta) ) / r^2 [/itex]

now I integrate ds from 0 to (pi * r ) / 2

so I get (pi*r)/2 after integrating

[itex] E = (*k*\Lambda * sin(theta) * pi * r ) / r^2 [/itex]

pi * r * lambda are = Q, right?

so now I have

[itex] E = kQ sin(theta) / r^2 [/itex]

The reason sin(theta) wasn't in my integration was because the change of the angle has no impact on the E right?

But sin(theta) = y / r

so now I'm left with

E = kQy / r^3But I have a feeling this is wrong. I can plug in for k, Q, and r^3, I'm not really sure what to do with y...
 
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  • #2
I think I'm wrong when I say that sin(theta) doesn't have an impact on E, because it does? Since the electric field is stronger as y is closer to 0?

So at the step

[itex] E = 2*k*\Lambda * sin(theta) / r^2 [/itex] , I find sin(theta) to be = y/r, so I get [itex] E = 2*k*\Lambda * ds * y / r^3 [/itex]

but how am I suppose to integrate y and ds ??
 
  • #3
Rijad Hadzic said:
1) Because I'm given L = .24 m, and the circumfrence of a circle is 2pir, since I have a semi circle I can set .24m = pi r, now r = .24m/pi
Yes.

The direction of the field is -j hat. Because the x components cancel out.
Yes

I consider a small piece of the rod I will call ds that has charge dQ. [itex] dQ = \Lambda ds [/itex] correct?
Yes, assuming ##\Lambda## is the linear charge density.

Anyways the magnitude of the electric field at a, E, is going to be 2dEsin(theta) because its getting its Y component from both sides of the Y axis.

Putting in the factor of 2 here is not a good idea. You are considering one element of charge dQ.

so

[itex] E = (2*k*\Lambda * ds * sin(theta) ) / r^2 [/itex]

The left side would be ##dE_y## and you could add a negative sign to the right side to indicate that the y component is negative. Again, I would drop the factor of 2 here.

now I integrate ds from 0 to (pi * r ) / 2
The arc length s and the angle θ are related. So, you should be able to write ds in terms of dθ and integrate with respect to θ. It's at this point that you can put in a factor of 2 and integrate over half the semicircle.
 
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  • #4
TSny said:
Yes.

The arc length s and the angle θ are related. So, you should be able to write ds in terms of dθ and integrate with respect to θ. It's at this point that you can put in a factor of 2 and integrate over half the semicircle.
ok so

[itex] dE = dE_y sin(thetha) [/itex]

[itex] dE_y = (k)(\Lambda) (ds ) / r^2 [/itex]

[itex] dE = (k)(\Lambda) (ds ) sin(theta) / r^2 [/itex]

since s = (r)(theta) and ds = rdThetha

[itex] E = (k)(\Lambda) (2)\int {sin(thetha) d(theta)} / r ^2 [/itex]

int sin thetha = -cos evaluate from 0 to pi/2 you get one thus

[itex] E = (k)(\Lambda)(2) / r^2 [/itex]

correct??

set lambda = Q/L so i have

[itex] E = (k)(\Lambda)(2) / (r^2) (L) [/itex]

I still feel like I am doing something wrong :(, am I?
 
  • #5
Rijad Hadzic said:
ok so

[itex] dE = dE_y sin(thetha) [/itex]
Shouldn't the symbols ##dE_y## and ##dE## be switched here?

since s = (r)(theta) and ds = rdThetha

[itex] E = (k)(\Lambda) (2)\int {sin(thetha) d(theta)} / r ^2 [/itex]
Looks like you forgot the r in rdθ.

int sin thetha = -cos evaluate from 0 to pi/2

you get one thus

[itex] E = (k)(\Lambda)(2) / r^2 [/itex]

correct??
You are off by a factor of ##r## on the right side.
set lambda = Q/L so i have

[itex] E = (k)(\Lambda)(2) / (r^2) (L) [/itex]
You can express L in terms of r and simplify. Should that be ##\Lambda## or ##Q## in your final result?
You almost have it.
 
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Related to Magnitude of the electric field

What is the magnitude of the electric field?

The magnitude of the electric field is a measure of the strength of the electric field at a specific point. It is represented by the symbol E and is measured in units of Newtons per Coulomb (N/C).

How is the magnitude of the electric field calculated?

The magnitude of the electric field is calculated by dividing the force exerted on a test charge by the magnitude of the test charge itself. This can be represented by the equation E = F / q, where E is the electric field, F is the force, and q is the test charge.

What factors affect the magnitude of the electric field?

The magnitude of the electric field is affected by the distance between charges, the magnitude of the charges, and the medium in which the charges are located. The electric field is stronger when the charges are closer together and weaker when they are farther apart. It is also stronger when the charges are larger in magnitude and weaker when they are smaller. The medium in which the charges are located can also impact the strength of the electric field.

How is the direction of the electric field determined?

The direction of the electric field is determined by the direction in which a positive test charge would move if placed in the field. The electric field lines point away from positive charges and towards negative charges.

What is the difference between electric field and electric field strength?

Electric field refers to the region in which an electric force is experienced, while electric field strength refers to the measure of the strength of the electric field at a specific point. Electric field strength is a scalar quantity, while electric field is a vector quantity that includes both magnitude and direction.

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