Electricity and Magnetism question. Please help ASAP!

Question

A uniformly charged insulating rod of length 17.0 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. The link to the figure is

https://app.gemoo.com/share/image-annotation/607459939978969088?codeId=DWaonZ6AwKAqL&origin=imageurlgenerator&card=607459939349823488

(a)

Find the magnitude of the electric field (in N/C) at O, the center of the semicircle.

b)

Find the direction of the electric field at O, the center of the semicircle. Pick one of the answers below

to the left

to the right    

upward

downward

into the page

out of the page

(c)

What if? What would be the magnitude of the electric field (in N/C) at O if the top half of the semicircle carried a total charge of −7.50 µC and the bottom half, insulated from the top half, carried a total charge of +7.50 µC?

(d)

What would be the direction of the electric field at O if the top half of the semicircle carried a total charge of −7.50 µC and the bottom half, insulated from the top half, carried a total charge of +7.50 µC? Pick one of the answers below

to the left

to the right    

upward

downward

into the page

out of the page

Answer 1

Hi Soyeb ,

1. When you bend the rod of length L into a semicircle, the radius of it, R , would be

R=L/π (1)

The linear density of charge σ on the semicircle would remain the same as of the rod:

σ=Q/L (2)

2. Imagine breaking the semicircle info tiny parts each with a length of dl . Each such part would carry a charge

dq=σdl=Q/L dl, (3)

and be at a radius R distance from the center O.

3. On the picture, the semicircle was shown to the left of the center so that the horizontal line going through the center is it's axis of symmetry. Let us draw two radius vectors R₁ and R₂ from the center O, to the start and to the end of the element dl . The difference between them,

|R₂-R|=|dR|=Rdθ would be equal to dl where dθ is the difference in angle that these radius vectors make to the horizontal.

4. According to the Coulomb Law, each element carrying charge

dq=σdl=(Q/L)Rdθ, (4)

would create an electric field of same absolute value |dE| but directed along the radius away from the center (b/c of Q<0).

Vectors dE would have horizontal and vertical components:

dE_ver=|dE|sinθ; dE_hor=|dE|cosθ (5)

where θ would change from (-π/2) at the lowest point of the semicircle to (+π/2) at the highest point.

5. From the symmetry of the setup we can see that for each element at an angle θ above the horizon there would be it's counterpart at the angle (-θ) below the horizon so that the vertical components of their electric field at point O would cancel each other [sin(-θ)=-sin(θ)]

6. As to the horizontal components, all elements would create them in the same direction : from the center O toward the semicircle along the horizontal.line.

7. Now, we could finally write the total field created by the semicircle as

E_v,tot= ∫(-π/2,+π/2)dE_hor=∫(-π/2,+π/2)|dE|cosθ=

∫(-π/2,+π/2){[k(Q/L)Rdθ]/R²}cosθ=

[k(Q/L)/R] ∫(-π/2,+π/2)dθcosθ=

[k(Q/L)/R]{sin θ|(-π/2,+π/2)}=

[k(Q/L)/R][-1-(-1)]=2[k(Q/L)/R] (6)

Replacing R=L/π in the last expression,

we get

E_v,tot=2πkQ/L^2 (7). where

k=is the Coulomb constant 8.99 x 10⁹Nm²/C²

Q=-7.50x10^-6C, is the change of the rod,

L=0.17m is the length of the rod

---------------

8. If the =-7.50x10^-6C semicircle is above and =+7.50x10^-6C semicircle is below horizontal, each would create the field

| E_v,tot|=2πk|Q|/L^2 (8)

that we just found, in the SAME (vertical) direction from the center straight up. So, the total field would be twice as strong:

| E_v,tot|=4πk|Q|/L^2 (9)

9. Finally, the last part of the problem seems repetitive.

I think what would make sense to ask here what if BOTH semicircles are charged with SAME charge (e.g. both with Q=-7.50x10^-6C

The answer here would be ZERO b/c both semicircles would create equal and opposite fields at the center O.

Hope this is helpful.

Best

Dr Ariel B.

ImaEvery part of a semicircle would create a

The key is to integrate the Coulomb Law