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3 years ago

11

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

distance "a" from the ring's center.

1 answer:

34kurt3 years ago

4 0

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

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Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un

elena55 [62]

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

7 0

4 years ago

At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three different directions. As a res

hram777 [196]

Answer:

F₃ = 122.88 N

θ₃ = 20.63°

Explanation:

First we find the components of F₁:

For x-component:

F₁ₓ = F₁ Cos θ₁

F₁ₓ = (50 N) Cos 60°

F₁ₓ = 25 N

For y-component:

F₁y = F₁ Sin θ₁

F₁y = (50 N) Sin 60°

F₁y = 43.3 N

Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:

F₂ₓ = F₂ = 90 N

F₂y = 0 N

Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:

F₁ₓ + F₂ₓ + F₃ₓ = 0 N

25 N + 90 N + F₃ₓ = 0 N

F₃ₓ = - 115 N

for y-components:

F₁y + F₂y + F₃y = 0 N

43.3 N + 0 N + F₃y = 0 N

F₃y = - 43.3 N

Now, the magnitude of F₃ can be found as:

F₃ = √F₃ₓ² + F₃y²

F₃ = √[(- 115 N)² + (- 43.3 N)²]

<u>F₃ = 122.88 N</u>

and the direction is given as:

θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)

<u>θ₃ = 20.63°</u>

7 0

3 years ago

In odd nuclei, what determines the final spin of the nucleus?

Katen [24]

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

$1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }$

Here,

$j=\frac{1}{2}$

and, L=1.

Fort parity,

$(-1)^{L}$

Put the value of L.

Parity will be -1.

Now, spin will be

$S=(\frac{1}{2} )^{-1}$ .

7 0

4 years ago

HELP!!! 30 POINTS+BRAINLIEST!!!!!QUICK!!

tia_tia [17]

Answer:

A:7.2 $ms^{-1}$

B:14.25 $ms^{-1}$

C:1.45 $sec$

D:10.3 $m$

E:2.9 $sec$

F:20.88 $m$

Explanation:

Let $v$ be the velocity and $\alpha$ be the angle between the velocity and ground.

Question A:

Horizontal component of velocity is given by $vcos(\alpha )$ .

So,horizontal component is $16\times cos(63)=16\times 0.45=7.2ms^{-1}$

Question B:

Vertical component of velocity is given by $vsin(\alpha )$ .

So,vertical component is $16\times sin(63)=16\times 0.89=14.25ms^{-1}$

Question C:

Time required is given by $\frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds$

Question D:

Maximum height is given by $\frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m$

Question E:

Time of flight is twice the time required to reach maximum height= $2\times 1.45=2.9 seconds$ .

Question F:

The distance between the player and ball after landing is called range and is given by $\text{horizontal component of velocity}\times \text{time of flight}=$ $7.2\times 2.9=20.88m$

8 0

3 years ago

008 (part 1 of 3) 10.0 points A 3.2 kg object is subjected to two forces, F~ 1 = (1.9 N) ˆı + (−1.9 N) ˆ and F~ 2 = (3.8 N) ˆı

musickatia [10]

Answer:

a' =4.15 m/s²

Explanation:

Given that

m= 3.2 kg

F₁ = 1.9 i −1.9 j N

F₂=3.8 i −10.1 j N

From second law of Newton's

F(net) = m a

F₁ + F₂ = m x a

1.9 i −1.9 j + 3.8 i −10.1 j = 3.2 a

a = 1.78 i - 3.75 j m/s²

The resultant acceleration a'

$a'=\sqrt{1.78^2+3.75^2}\ m/s^2$

a' =4.15 m/s²

6 0

4 years ago