# Calculating eletric potential using line integral of electric field

• pedromatias
In summary, the conversation discusses the calculation of electric potential using different methods. The first method involves calculating the electric field at a point and then using it to find the potential through integration. However, this method does not give the correct result. The second method involves directly integrating the expression for electric field, which gives the correct solution. The conversation concludes with a discussion on verifying the results obtained from a calculator.
pedromatias
Homework Statement
A bar of length l has charge Q that is linearly distributed, with constant charge density λ. There is a point P at distance D from the center of mass of the bar, and P is in the same axis as the bar. Knowing the electric field, calculate the electric potential at P using a line integral. See figure.
Relevant Equations
The difference of potential between two points is symmetric to the line integral between those two points of the dot product between the electric field and the line.
So, I am able to calculate the electric potential in another way but I know that this way is supposed to work as well, but I don't get the correct result.

I calculated the electric field at P in the previous exercise and its absolute value is $$E = \frac {k Q} {D^2-0.25*l^2}$$ This is correct as per the solution. From the figure, we can see that the direction of the electric field at P is that of the negative x direction.

So, to calculate the electric potential I took advantage of the fact that at infinite the potential is zero and did

$$V_P = \int_D^{\infty} \frac {k Q} {r^2-0.25*l^2} \, dr$$

However, this integral does not give the correct result. I inserted it into integral calculator and the result was not the correct one, which is

$$k \lambda \ln \frac {D+\frac{l}{2}} {D-\frac{l}{2}}$$

I don't understand what I am doing incorrectly. My reasoning is that I have the expression for the Electric field in a point at a distance ##r## in the axis of ##P##, and that Electric field points in the same direction as the straight line path from ##P## to ##\infty##, so the dot product is the magnitude of the electric field.

The way I solved it correctly and which gives me the right solution is, where ##x## is the distance from P,

$$dv = k \frac {dq}{x}$$
$$dv = k \lambda \frac {dx}{x}$$
$$V = \int_{D-0.5l}^{D+0.5l} \frac {k \lambda} {x} \, dx$$
$$V = k \lambda \ln \frac {D+\frac{l}{2}} {D-\frac{l}{2}}$$

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What result does your calculator give? How have you verified that it is different from the one you got? Sometimes solutions given by calculators look different but aren't.

pedromatias and vela

## 1. What is the concept of electric potential?

Electric potential is the amount of electric potential energy per unit charge at a specific point in an electric field.

## 2. How is electric potential calculated using line integral of electric field?

To calculate electric potential using line integral of electric field, we use the formula V = -∫E•dl, where V is the electric potential, E is the electric field, and dl is the differential length along the path of integration.

## 3. What is the significance of using line integral in calculating electric potential?

Line integral allows us to take into account the direction and magnitude of the electric field at every point along the path of integration, providing a more accurate calculation of electric potential.

## 4. Can electric potential be negative?

Yes, electric potential can be negative. This indicates that the electric field is directed in the opposite direction of the path of integration.

## 5. How is electric potential related to electric potential energy?

Electric potential energy is the potential energy that a charged particle possesses due to its position in an electric field. The electric potential is the measure of this potential energy per unit charge at a specific point in the electric field.

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