Question

Number 10 plz

Aundornly charged insulating rod of length 20.0㎝i, bent into the shape ofasemade as shown n the figure below. The red hauteldige eende (a) Find the magnitude of the electric flield in N/C) at O, the center of the semicircle. N/C (b) Find the direction of the electric lelid at O, the center of the semicircle to the leh downwarts MacBook Air

Answer 1

10.

a)

Q = total charge on the rod = - 7.50 x 10^-6 C

L = length of the rod = 20 cm = 0.20 m

Charge per unit length is given as

$\lambda$ = Q/L = (- 7.50 x 10^-6)/0.20 = - 37.5 x 10^-6 C/m

r = radius of the semicircle

Length of rod = circumference of the semicircle

L = 2 $\pi$ r

0.20 = 2 (3.14) r

r = 0.032 m

consider a small length "dl" at angle "$\theta$"

dq = small charge on the small length = $\lambda$ dl = $\lambda$ r d$\theta$

small electric field at "O" due to small length "dl" is given as

dE = k dq/r²

dE = k $\lambda$ r d$\theta$/r²

dE = k $\lambda$ d$\theta$/r

The component of electric field along the Y-direction cancels out and we get net electric field along the x-direction as

E_x = $\int$dE Cos$\theta$

E_x = $\int_{0}^{\pi }$ (k $\lambda$ Cos$\theta$d$\theta$/r)

E_x = 2 k $\lambda$/r

inserting the values

E_x = 2 (9 x 10⁹) (37.5 x 10^-6)/(0.032)

E_x = 2.1 x 10⁷ N/C

b)

direction : to the left