Question

42
A non-conducting rod of length 2.2 m carries a negative charge of 3.8 * 10-C spread uniformly
over its length. What is the magnitude of electric field near the mid point of the rod at a
perpendicular distance 3.6 mm from the rod?
​

Answer 1

Answer:  Explanation: Below given hope u understand

Answer 2

Hello dear,

◆ Answer -

E = 8.65×10^5 N/C

● Explanation -

# Given -

Q = 3.8×10^-7 C

l = 2.2 m

r = 3.6 mm = 3.6×10^-3 m

# Solution -

Linear charge density of the conductor is given by -

λ = Q / l

λ = 3.8×10^-7 / 2.2

λ = 1.73×10^-7 C/m

Electric field on the perpendicular mid-axis of rod is calculated by -

E = (1/4πε0) 2λ/r

E = (9×10^9 × 2 × 1.73×10^-7) / (3.6×10^-3)

E = 8.65×10^5 N/C

Therefore, the magnitude of electric field near the mid point of the rod at a

perpendicular distance 3.6 mm from the rod is 8.65×10^5 N/C.

Thanks for asking..