Ask question

Ask question

All categories

4 years ago

9

A uniformly charged ring of radius 10.0 cm has a total charge of 66.0 μC. Find the electric field on the axis of the ring at the

following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.) (a) 1.00 cm

1 answer:

Tems11 [23]4 years ago

5 0

To solve this problem we will apply the mathematical consideration of the electric field on an axial axis of a ring. This definition is already established mathematically and is a subordinate of the definition of the magnetic field of Coulomb's laws. It can be expressed as,

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

Here,

k = Coulomb's constant

Q = Charge

x = Distance on the axial line

R = Radius of the circle or ring

At x = 1 cm, we have that the total charge is 66.0 μC and the radius 0.1m, then replacing,

$E = \frac{(9*10^9)(66*10^{-6})(0.01)}{\sqrt{((0.01)^2+(0.1)^2)^3}}$

$E = 5.85*10^6 N/C$

Therefore the electric field on the axis of the ring is 5.85MN/C

You might be interested in

What type of rays can pass through all matter?

Semmy [17]

Gamma rays can pass through a lot

7 0

3 years ago

A(n) ________ describes the shape of a planet’s orbital path. astronomical unit

sleet_krkn [62]

It is an ellipse.

Hope this helps!

8 0

3 years ago

Read 2 more answers

I need help plz and thank you. This is past due and I will mark brainiest

Zielflug [23.3K]

Answer:

reproduced in print form for non-profit educational use with. Reading ... 2. I had them in my backpack. 3. Help me look for them. 4. We don't want to get lost. 5. I'm so ... Commas set off introductory words, such as yes, no, and thank you. Rewrite ... Correct each run-on sentence below by writing it as a compound sentence. 1.

Explanation:

4 0

3 years ago

Read 2 more answers

A 3 kg object is moving along a horizontal surface. The kinetic energy of the object is increasing at a constant rate of 6 J/m;

Whitepunk [10]

To solve this problem we will apply the concepts of energy conservation and Newton's second law that defines force as the product of the object's mass with its acceleration. Additionally we will apply concepts related to the kinematics equations of linear motion.

For conservation of energy we have that work is equal to kinetic energy therefore,

$W = KE$

$Fd = \frac{1}{2} mv^2$

Here,

F = Force

d = Displacement

m = Mass

v= Velocity

At the same time we have the relation of

$F = \frac{W}{d}$

Therefore the value of the force can be interpreted as the rate of increase in energy per unit of distance, which makes it equivalent to

$F = \frac{W}{d} = 6J/m$

Applying Newton's Second Law

$F = ma$

$6J/m = (3kg)a$

$a = 2m/s^2$

In 4 seconds final velocity of the object becomes

$v = at$

$v= 2*4$

$v= 8m/s$

Then the work done is equal to,

$W = KE$

$W = \frac{1}{2} mv^2$

$W = \frac{1}{2} (3)(82)$

$W = 96J$

Then the displacement is,

$W = F*d$

$d = \frac{W}{F}$

$d = \frac{96}{6}$

$d = 16 m$

Therefore the distance moved is 16m

7 0

4 years ago

A plasma is a gas of ionized (charged) particles. When plasma is in motion, magnetic effects "squeeze" its volume, inducing inwa

zzz [600]

Answer:

a

The volume charge density is $\rho = nq$

b

The surface charge density is $\sigma = n^{\frac{2}{3} } q$

Explanation:

From the question we are told that

The radius is R

The length is L

The velocity is v

The number of ions per unit volume is n

The charge is q

The thickness of the cylinder surface is $n^{\frac{1}{3} }$

The volume charge density is mathematically represented as

$\rho = nq$

The surface charge density is mathematically represented as

$\sigma = \rho n^{\frac{1}{3} }$

substituting for $\rho$

$\sigma = n * n^{\frac{1}{3} } q$

$\sigma = n^{\frac{2}{3} } q$

5 0

3 years ago