a) What is the magnitude of the electric field at point P , which is on the posi

Question

a)

What is the magnitude of the electric field at point P, which is on the positive x-axis at x = 44.0 cm ?

b)

What is the direction of the electric field at point P?

+/-

c)

A particle with a charge of 2.10 C is placed at the point P described in part A. What is the magnitude of the force exerted by the particle on the ring?

d)

What is the direction of the force exerted by the particle on the ring?

+ or -

A ring-shaped conductor with radius a 2.50 cm has a total positive charge Q J 0.122 nC uniformly distributed around it. (Eigure 1) Figure 1 v of 1 ds dE dE

Explanation / Answer

(a) the eletric field components along y axis ()sin components) cancel and total field = sum of horizontal components . the electric field on the axis is given by

E= kqx/(x2 + a2)3/2 along x axis

E= (8.99x10^9)(0.122x10^-9)(0.44)/(0.442 + 0.0252)3/2

. E= 5.64 N/c .

(b)the direction of the electric field is along the positive x axis (away from the ring) so +

(c) q= 2.1 uC= 2.1x10^-6 c

force= f= qE

f= 2.1x10^-6 (5.64)

f=1.1844 x10^-5 N .

(d) the direction of force on the ring is along the positive x axis. so +

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