Electric field strength at the center of semicircle

The radius of the semicircle should be 0.06 m, not 0.12 m. Also, the negative sign should be dropped since the magnitude of a vector is always positive.In summary, the electric field at the center of a semicircular rod with a total charge of -7.50 µC is 11936620 N/C, calculated using the equation E = 2kq / pi r2 with a radius of 0.06 m. The negative sign should be dropped since the magnitude of the electric field is always positive.
  • #1
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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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  • #2
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 ...?
 
  • #3
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...?
 
  • #5
You used the wrong radius.
 

FAQ: Electric field strength at the center of semicircle

1. What is the formula for calculating electric field strength at the center of a semicircle?

The formula for calculating electric field strength at the center of a semicircle is E = kQ/R, where E is the electric field strength, k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the charge of the semicircle, and R is the radius of the semicircle.

2. How does the electric field strength at the center of a semicircle vary with changes in the charge and radius?

The electric field strength at the center of a semicircle is directly proportional to the charge and inversely proportional to the radius. This means that as the charge increases, the electric field strength increases, and as the radius increases, the electric field strength decreases.

3. What is the direction of the electric field at the center of a semicircle?

The direction of the electric field at the center of a semicircle is perpendicular to the surface of the semicircle, pointing away from the center if the charge is positive and towards the center if the charge is negative. This is because electric field lines always point away from positive charges and towards negative charges.

4. How does the electric field at the center of a semicircle compare to that of a full circle with the same charge and radius?

The electric field at the center of a semicircle is half the value of the electric field at the center of a full circle with the same charge and radius. This is because the electric field at the center of a full circle is evenly distributed over the entire circumference, while the electric field at the center of a semicircle is only distributed over half the circumference.

5. What are some real-world applications of the concept of electric field strength at the center of a semicircle?

The concept of electric field strength at the center of a semicircle is important in understanding the behavior of electric fields in various electronic devices, such as antennas, sensors, and capacitors. It is also relevant in the study of electromagnetism and electric circuits, and has applications in industries such as telecommunications and power generation.

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