E-field problem(integration)

• kenok1216
In summary: No, integrating again won't help. The force on the rod is dependent on the distance from the point charge, not the charge itself.

kq/r^2=E
λ=q/L

The Attempt at a Solution

λ=8μC/1=8μC/m
dq=λdx
dE=k(dq)/(√(x^2+2^2))^2
dE=kλdx/(x^2+4)
-----------------------------------------------
for x-direction
dEx=dEcosθ=(2kλdx)/(x^2+4)^(3/2)
Ex
=2kλ∫dx/(x^2+4)^(3/2)
=2kx[x/4(x^2+4)^(1/2)] (from 0 to 1)
=2kλ/4√5=16099.68V/m
----------------------------------------------------
for y-direction
dEy=-dEsinθ=-kλxdx/(x^2+4)^(3/2)
Ey
=-kλ∫xdx/(x^2+4)^(1/2) (from 0 to 1)
=-kλ(-1/√5+0.5)=-3800.6V/m
------------------------------------------------------------
Enet=√[(Ex)^2+(Ey)^2]=16.54kV/m
diretion=tanEy/Ex=-13.28degree
----------------------------------------------------------
but the answer of (b)(i) E=24.65KV/m diretion=-13.28degree
what is wrong during the calculation ?
also no idea on (b)(ii)

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kenok1216 said:

Homework Statement

View attachment 99562

kq/r^2=E
λ=q/L

The Attempt at a Solution

λ=8μC/1=8μC/m
dq=λdx
dE=k(dq)/(√(x^2+2^2))^2
dE=kλdx/(x^2+4)
-----------------------------------------------
for x-direction
dEx=dEcosθ=(2kλdx)/(x^2+4)^(3/2)
Ex
=2kλ∫dx/(x^2+4)^(3/2)
=2kx[x/4(x^2+4)^(1/2)] (from 0 to 1)
=2kλ/4√5=16099.68V/m
----------------------------------------------------
for y-direction
dEy=-dEsinθ=-kλxdx/(x^2+4)^(3/2)
Ey
=-kλ∫xdx/(x^2+4)^(1/2) (from 0 to 1)
=-kλ(-1/√5+0.5)=-3800.6V/m
------------------------------------------------------------
Enet=√[(Ex)^2+(Ey)^2]=16.54kV/m
diretion=tanEy/Ex=-13.28degree
----------------------------------------------------------
but the answer of (b)(i) E=24.65KV/m diretion=-13.28degree
what is wrong during the calculation ?
also no idea on (b)(ii)
the point charge of -3μC will affect the result of Ex?

kenok1216 said:
the point charge of -3μC will affect the result of Ex?
anyone help...

kenok1216 said:
the point charge of -3μC will affect the result of Ex?

No it wont, your method seems fine to get the Electric field assuming the values plugged in and integration done was right. For the second part, F=qE can you make something of it?

my handwrite version, something wrong in here?
for the point charge act on the rod, how can i cal the e-field ,do integration again??
AbhinavJ said:
No it wont

AbhinavJ said:
No it wont, your method seems fine to get the Electric field assuming the values plugged in and integration done was right. For the second part, F=qE can you make something of it?
but the answer of 12(b)(i) is
24.65 kVm-1;
-13.28°

Yes for the point charge on rod, you'll have to Integrate again. dF=lamdadx*E, where E is the Electric field at that particular point. Similar to the above method. And maybe the answer for the 1st question in the book could be wrong, because the angle is coming out right that meand Ex and Ey are correct or youre incredibly lucky.

AbhinavJ said:
Yes for the point charge on rod, you'll have to Integrate again. dF=lamdadx*E, where E is the Electric field at that particular point. Similar to the above method. And maybe the answer for the 1st question in the book could be wrong, because the angle is coming out right that meand Ex and Ey are correct or youre incredibly lucky.
q = 8μC?(charge of rod object A) how dq related to dx? dq=λdx?? λ of point charge?? -3μC/0?? since part 2 is want to calculate force on object A, so λ is the λ of the point charge?

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kenok,
I think your answer for part (i) in the OP is probably correct. I have not checked the numbers carefully. But you can easily check that the answer of 24.65 kV/m cannot be correct. Can you see that even if you concentrated all of the charge of the rod as a point charge at the base of the rod, it would not produce that much electric field at B.

For part (ii), is there any relation between the force felt by the rod and the force felt by the charge at B?

TSny said:
kenok,
I think your answer for part (i) in the OP is probably correct. I have not checked the numbers carefully. But you can easily check that the answer of 24.65 kV/m cannot be correct. Can you see that even if you concentrated all of the charge of the rod as a point charge at the base of the rod, it would not produce that much electric field at B.

For part (ii), is there any relation between the force felt by the rod and the force felt by the charge at B?
action and reaction pair so Fb=-Fa?

kenok1216 said:
View attachment 99599
q = 8μC?(charge of rod object A) how dq related to dx? dq=λdx?? λ of point charge?? -3μC/0?? since part 2 is want to calculate force on object A, so λ is the λ of
the point charge?

Each part of the rod is at a different distance from the point charge, thereby feels a different force. Get force as a function of x, integrate.

AbhinavJ said:
Each part of the rod is at a different distance from the point charge, thereby feels a different force. Get force as a function of x, integrate.
thank for you , but final i do not do integrate in part 2
using Fa=-Fb can solve this problem more faster since Fb=qEb and Eb is calculate at part 1

kenok1216 said:
using Fa=-Fb can solve this problem more faster since Fb=qEb and Eb is calculate at part 1
Yes, you can do that.