Window Help PHYS4BSP17 Exercise 21.53 Exercise 21.53 Aring-shaped conductor with

Question

Window Help PHYS4BSP17 Exercise 21.53 Exercise 21.53 Aring-shaped conductor with radius a 220 cm has a total positive charge Q 0.128 nC uniformly distributed (Eigure 1) Figure 1 of 1 16% El Sun 6:30 PM Ahmad Yousef session masteringphysics.com signed in as Ahmad Yousef belo previous l 13 of 15 l next Part A What is the magnitude ofthe electric field at point P. which is on the positive xaxis at a 45.0 E. 7.22 N/C Submit My Answers Give Up Incorrect Try Again Part B What is the direction of the electric field at point P? +x-direction My Answers Give Up Correct

Explanation / Answer

radius , a = 2.2 cm = 0.022 m

Q = 0.128 nC

part A)

magnitude of electric field at P = k * Q * x/(a^2 + x^2)^1.5

magnitude of electric field at P = 9 *10^9 * 0.128 * 10^-9 * 0.45/(0.022^2 + 0.45^2)^1.50

magnitude of electric field at P = 5.67 N/C

part B)

the direction of electric field at P is tot he right

the electric field is in + x direction

part c)

q = -3 uC

magnitude of force exerted by the particle on the ring = q * E

magnitude of force exerted by the particle on the ring = 3 *10^-6 * 5.67

magnitude of force exerted by the particle on the ring = 1.702 *10^-5 N

part d)

the direction of force on the ring is + x direction

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