Electric field strength at the center of semicircle

AI Thread Summary
To find the electric field strength at the center of a semicircular rod with a total charge of -7.50 µC, the initial calculation used an incorrect approach, leading to an erroneous magnitude of 11,936,620 N/C. The contributions to the electric field from different segments of the rod point in various directions, resulting in a net vector that is smaller than calculated. The correct method requires considering the vector nature of electric fields and the geometry of the semicircle. Additionally, the radius used in the calculation was incorrect, which affected the results. Accurate references such as hyperphysics or Wikipedia can provide further guidance on electric field calculations.
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Homework Statement


A uniformly charged insulating rod of length 12.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of −7.50 µC.

Find the magnitude of the electric field at O, the centre of the semicircle.

Homework Equations


E = 2kq / pi r2
(Acquired on the internet)

The Attempt at a Solution


E = 2(9X109)(-7.5X10-6) / pi (0.06)2
E = -135000 / 0.01130973355
E = -11936620.73

magnitude: 11936620 N/C

Got it wrong... I don't know why...
 
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You want the magnitude of a vector ##\vec E##. Contributions from different places on the rod point in different directions, so the sum vector is not as big as you calculate now.
If the ring were not half but complete, ##|\vec E|## would be ...?
 
BvU said:
You want the magnitude of a vector ##\vec E##. Contributions from different places on the rod point in different directions, so the sum vector is not as big as you calculate now.
If the ring were not half but complete, ##|\vec E|## would be ...?

Less...?
 
You used the wrong radius.
 
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