Question

A semicircular ring of radius 0.5m is uniformly charged with a total charge of 1.4*10^9 C the electric field intensity at the center of the ring is

Answer 1

electric field intensity at the centre of ring of radius r is given by, $E=\frac{2k\lambda}{r}$

where $\lambda$ is charge per unit length or linear charge density of semicircular ring.

e.g., $\lambda$ = q/πr [ perimeter of semicircle = πr ]

= 1.4 × 10^-9/(3.14 × 0.5) [ total charge given , 1.4 × 10^-9 C ]

= 2.8/(3.14) × 10^-9 C/m

now, electric field intensity , E = 2 × 9 × 10^9 × (2.8/3.14) × 10^-9/(0.5)

= 18 × 5.6 /3.14

= 32.10 N/C ≈ 32 N/C

hence, answer is 32 N/C

Answer 2

Answer:

Accordingtoquestion:

r=0.5m

q=1.4×10

−9

C

Now,findelectrifieldintensity:

θ=

radius

arc

⇒dl=Rdθ−−−−−(i)

let,cheofdensity=p

cheofthiselements=dq

⇒dq=Rdθ−−−−−−−−(ii)

Now,

Eletricfieldduetotheelementsatlater:

dE(magnitude)=Kdq/R

2

dE=Kdq/R

2

cosθi+Kdq/R

2

sinθ(−j

dE=

R

2

Kdq

[cosθi−sinθj]

E=

R

Kp

[−2j]

Now,

wehaveP(chargeperunitlength)=

πR

θ

sothat,

E=

πR

2

2K

(−j)

E=

π×(0.5)

2

2×9×10

9

×1.4×10

−9

∴E=32V/m