Potential Difference Near a Charged Rod

• lightupshoes8
In summary, we have a long rod with a uniform charge and we need to calculate the potential difference between two points, C and A, located at different distances from the rod. By setting up an integral and taking into account the distance a, we can find that the potential difference will be equal to the negative of the product of a constant and the integral of 1 over the distance from B to C.
lightupshoes8

Homework Statement

A long rod of length L = 7.8 m carries a uniform charge Q = -6.5e-08 coulombs.
Calculate the potential difference VC - VA. The distance c is 4 cm, the distance b is 11 cm, and the distance a is 4 cm. Only part of the rod is shown in the diagram (it is longer than shown).

Homework Equations

Erod = (9e9)*$$\frac{2Q/L}{r}$$
$$\Delta$$V = - $$\int$$ E dl

The Attempt at a Solution

I figured that since A is the same distance horizontally as B from the rod, there would be no change in potential since its not acting in that direction. So I was just going to find Vc-Vb

I set up my integral as follows:

V = - (9e9)*$$\frac{2Q}{L}$$$$\int$$ $$\frac{dr}{r}$$
Integral going from upper to lower: B to C

I just don't think I'm headed in the right direction. It seems like the answer is not going to be what I want it to be.

Last edited:
lightupshoes8 said:
I figured that since A is the same distance horizontally as B from the rod, there would be no change in potential since its not acting in that direction. So I was just going to find Vc-Vb
That's right.
I set up my integral as follows:

V = - (9e9)*$$\frac{2Q}{L}$$$$\int$$ $$\frac{dr}{r}$$
Integral going from upper to lower: B to C
That looks ok too.

As a scientist, my response to this content would be as follows:

The potential difference near a charged rod can be calculated using the formula V = - ∫E dl, where E is the electric field and dl is the infinitesimal distance along the path of integration. In this case, the rod has a uniform charge of Q = -6.5e-08 coulombs and a length of L = 7.8 m.

To calculate the potential difference between points C and A, we need to integrate the electric field along the path from point B to C, as the electric field is perpendicular to the path from A to B. The electric field at a distance r from the rod is given by E = (9e9)*\frac{2Q/L}{r}.

Therefore, the potential difference between points C and A can be calculated as follows:

ΔV = - ∫E dl = - (9e9)*\frac{2Q}{L} ∫\frac{dr}{r} = - (9e9)*\frac{2Q}{L} ln\frac{r}{r0} (from B to C)

Where r0 is a reference distance, in this case, it can be taken as the distance from the rod to point B. Substituting the values given in the problem, we get:

ΔV = - (9e9)*\frac{2(-6.5e-08)}{7.8} ln\frac{0.04}{0.11} = 0.0049 volts

Therefore, the potential difference between points C and A is 0.0049 volts. It is important to note that the potential difference is independent of the reference point chosen, as long as it is outside the region of the electric field.

What is potential difference near a charged rod?

Potential difference, also known as voltage, is the difference in electric potential energy per unit of charge between two points near a charged rod. It is a measure of the strength of the electric field created by the charged rod.

How is potential difference near a charged rod calculated?

Potential difference is calculated by dividing the electric potential energy by the amount of charge. In the case of a charged rod, the electric potential energy is the product of the charge on the rod and the electric field created by the rod.

What factors affect the potential difference near a charged rod?

The potential difference near a charged rod is affected by the amount of charge on the rod, the distance from the rod, and the medium surrounding the rod. The potential difference is stronger when the charge is larger, the distance is smaller, and the medium is a good conductor.

What is the relationship between potential difference and electric field near a charged rod?

The electric field near a charged rod is directly proportional to the potential difference. This means that as the potential difference increases, the electric field also increases. In other words, the stronger the potential difference, the stronger the electric field near the rod.

How does potential difference near a charged rod affect the movement of charges?

Charged particles will experience a force when placed near a charged rod due to the potential difference. The direction of the force will depend on the charge of the particles and the direction of the electric field. If the potential difference is high, the force will be stronger, causing the particles to move more rapidly.

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