# Work required to move a charged ball

In summary, a ring of diameter 7.10 cm is fixed in place and carries a charge of 5.00 μC uniformly spread over its circumference. It takes work to move a tiny 3.40 μC charged ball of mass 1.70 g from very far away to the center of the ring.

## Homework Statement

A ring of diameter 7.10 cm is fixed in place and carries a charge of 5.00 μC uniformly spread over its circumference. How much work does it take to move a tiny 3.40 μC charged ball of mass 1.70 g from very far away to the center of the ring?

## Homework Equations

I tried using: $$W=-\Delta U$$

## The Attempt at a Solution

We know that when the ball is very far away U=0J

Then to find the potential energy when the ball is in the center of the ring I did:
Change in charge is $$5.00X10^{-9}C/2\pi(0.0710m)=1.12X10^{-8}C/m$$

$$\int \frac{k*q*dq}{r} ds$$ from 0 to 2pi.

$$\frac{k*q*dq}{r} \int ds$$ from 0 to 2pi.

$$\frac{8.988X10^{9}N*m^2/C^2*3.40X10^{-9}C*1.12*10^{-8}C/m}{0.0710m} \int ds$$ from 0 to 2pi $$=3.031X10^-5J$$.

$$W=-\Delta U=-(3.031X10^-5J-0J)=-3.031X10^-5J$$

I'm having trouble following your integral from the first line. I'm thinking V = kq/r for a point charge so dV = k/r*dq, where r is the radius of the ring.
Ah, you are doing energy E = QV so dE = QdV = kQ/r*dq
I'm using q for the charge on the ring, Q for the charge on the particle.
Your s must be something to do with the arc length going round the ring.
So dq = q/(2*pi*r)*ds
Using s = r*A where A is the angle running from 0 to 2pi, you get ds = r*dA and
dq = q/(2*pi)dA
so the integral is dE = kQq/(2*pi*r)*dA from A = 0 to 2*pi.
integral 0 to 2*pi of dA is just 2*pi.
E = kQq/r.
Well, I guess we didn't need to integrate at all! Because all the charge is the same distance from the center point.

I'm still slightly confused. So I can find E at the center of the ring using E=kQq/r. But, since W=-(U_f-U_i) how do I get U from E? Since E=kq/r^2 and U=kq_1*q_2/r would you just multiply the E you get by qr and if so what q would you use: the charge on the ring or the charge on the ball? Thanks for your help.

Try this approach. Find E as a function of r along the axis of the ring using integration. Then

$$V=-\int_{\infty}^{r}\mbox{Edr}$$

The work done is

$$W=Vq$$

Is the mass in a gravitational field also and if so what is the orientation of the ring axis with the field?

## 1. What is work required to move a charged ball?

The work required to move a charged ball refers to the amount of energy needed to move the ball from one point to another against an electric field. This work is typically measured in units of joules (J).

## 2. How is the work required to move a charged ball calculated?

The work required to move a charged ball is calculated using the formula W = qV, where W is the work, q is the charge of the ball, and V is the potential difference or voltage between the two points. This formula is based on the concept of electric potential energy.

## 3. What factors affect the work required to move a charged ball?

The main factors that affect the work required to move a charged ball are the amount of charge on the ball, the distance between the two points, and the strength of the electric field. The work required will increase as the charge or distance increases, and will decrease as the strength of the electric field increases.

## 4. Can the work required to move a charged ball be negative?

Yes, the work required to move a charged ball can be negative if the ball is moving in the direction of the electric field. This means that the electric field is doing work on the ball, rather than the ball doing work against the electric field. In this case, the work is considered to be negative because the electric field is providing the energy for the movement of the ball.

## 5. What are some real-life applications of work required to move a charged ball?

The concept of work required to move a charged ball has many practical applications, such as in the design of electrical circuits and devices. It is also important in understanding the behavior of charged particles in electric fields, which is crucial in fields such as particle physics and astronomy. Additionally, the concept is used in the development of technologies such as batteries and solar cells.

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