3 A ring shaped conductor with radius a - 2.50 cm has a total positive charge Q

Question

3 A ring shaped conductor with radius a - 2.50 cm has a total positive charge Q +0.125 nC uniformly distributed around it. (See figure). The center of the ring is at the origin of coordinates O what is magnitude and direction of the electric field at point P, which is on the x-axis at x 40.0 em? A proton is placed in a uniform electric field and then released. Then an electron is placed at this same point and released. Do these two particles experience the same force? The sime acceleration? a. b-

Explanation / Answer

A )Assuming the ring of total charge Q, is made up of charge elements,consider an element of length dl and charge dq

As charge Q is uniformly spread on ring of radius 'a', charge on 1 unit length is [ Q /2(pi) a ]

Charge on any element of length dl =dq =[ Q/2(pi)a ]dl

Electric field due to an element,at a point on the axis =dE = kdq/d^2

Where d = sq rt [x^2+a^2]

Relving dE into components,

component along the axis =dEcosO=dE(x/d)

component perpendicular to the axis =dEsinO=dE(a/d)

Considering a pair of diametrically opposite elements,the components perpendicular to the axis cancel out and components along the axis are added up.

Contribution along the axis due to a pair =2dE(x/d)

Contribution along the axis due to one element = dE(x/d)

Contribution along the axis due to the entire ring is found by summation (integration) over entire ring.

Electric field due to ring =E = summation ( kdq/d^2 )(x/d)

Substituting , dq =( Q/2pia )dl

E = summation[ k( Q/2pia)*dl*x/d^3 ]

E = k (Q/2pia)*(x/d^3)summation 'dl'

Substituting , summation dl =2(pi)a and d = sq rt [x^2+a^2]

E = k Q*x / ( x^2+a^2 )^3/2

x = 40 cm =0.40 m

radius = a = 2.50 cm = 0.025 m

total positive charge Q= 0.125nC =1.25*10^-10 C

E = [9*10^9] *(1.25*10^-10 )*0.40 / ( 0.40^2+0.025^2 )^3/2

E = 0.45 / ( 0.104 )

E =4.326 N/C

The magnitude of the electric field at point P at x = 40.0 cm is 4.326 N/C

The direction of the electric field at point P is +x-direction

B)

the force experienced is same in magnitude but in opposite direction.And as the magnitude of force is same, the acceleration is also same in magnitude.but here proton and electron have different masses. so the acceleration is different.for your information: F = m*a. since F is same. the particle with less mass have more acceleration and in this case electron has less mass. so the acceleration of electron is more than that of proton

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.