Electric field, direction and magnitude

AI Thread Summary
The discussion revolves around calculating the electric field due to a uniformly charged rod at a specific point. The user initially calculated the electric field magnitude as 6.17 * 10^5 N/C, while the correct answer is 1.59 * 10^6 N/C. A key point raised is the importance of considering the direction of the electric field as a vector, which may have been overlooked in the calculations. The user is advised to refer to textbooks for clarification on the electric field due to a wire and to ensure the correct interpretation of the axis of the rod. Accurate calculations require careful consideration of both magnitude and direction.
minifhncc
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Homework Statement



A rod 14.0 cm long is uniformly charged and has a total charge of -22.0μC. Determine
(a) the magnitude and
(b) direction of the electric field along the axis of the rod at a point 36.0 cm from its centre.

[I can do (b)]

Homework Equations



Electric field at the point due to a small element, ΔE=(k Δq)/(x^2)

The Attempt at a Solution


I got down to the total field at P being,
(kQ/l) * integral of 1/x^2 from x=0.29 and x=0.43

and I get 6.17 * 10^5 NC^-1

But the answer says 1.59 * 10^6 NC^-1

Are the answers wrong?
 
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The given answer is right, how did you get that wrong value?

ehild
 
ehild said:
how did you get that wrong value?

(kQ/l) * integral of 1/x^2 from x=0.29 and x=0.43

= 8.9876* 10^9 * -22 * 10^-6 / 0.14 ... etc
 
minifhncc said:
(kQ/l) * integral of 1/x^2 from x=0.29 and x=0.43

= 8.9876* 10^9 * -22 * 10^-6 / 0.14 ... etc

And what is in place of etc?

ehild
 
I don't think you have considered the direction of E.
E is a vector, and you need to treat the two components separately.
Check any book for the field due to a wire, and you'll understand what you've missed.
 
It depends what is meant on "axis of the rod". It should be the straight line along the rod.

ehild
 
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