I need help with the following problem:

A charged rod, uniformly charged with ‘Q’, with length ‘2a‘ is in vacuum. Find vector of electric field at point ‘A(0,a)‘ and electric potential at point ‘B(0,0)‘ with referent point at infinity.

Thanks for replies.

To solve the problem of finding the electric field at point A(0, a) and the electric potential at point B(0, 0) due to a uniformly charged rod of length 2a, we can break it down into manageable steps. Let's start with the electric field calculation.

Electric Field at Point A(0, a)

We have a uniformly charged rod of length 2a, which means it extends from -a to +a along the x-axis. The total charge on the rod is Q. To find the electric field at point A(0, a), we can use the principle of superposition, which states that the total electric field is the vector sum of the electric fields due to each infinitesimal charge element on the rod.

Setting Up the Problem

Let’s denote an infinitesimal charge element on the rod as dq. The linear charge density λ can be defined as:

• λ = Q / (2a)

Now, we can express dq in terms of λ:

• dq = λ dx = (Q / (2a)) dx

Next, we need to find the electric field contribution dE at point A due to this charge element dq. The distance from the charge element located at position x to point A is:

• r = √(x² + a²)

The electric field due to dq at point A is given by Coulomb's law:

• dE = (1 / (4πε₀)) * (dq / r²)

Calculating the Components

Since the electric field is a vector, we need to consider its components. The vertical component (along the y-axis) will be:

• dEy = dE * (a / r)

And the horizontal component (along the x-axis) will cancel out due to symmetry, as for every charge element at position x, there is an equal charge element at position -x. Thus, we only need to integrate the vertical component:

Integrating Over the Rod

The total electric field at point A can be found by integrating dEy from -a to +a:

• E = ∫(dEy) = ∫[ (1 / (4πε₀)) * (λ dx * (a / (x² + a²)^(3/2))) ] from -a to +a

Substituting λ into the integral gives:

• E = (Q / (8πε₀a)) * ∫[ (1 / (x² + a²)^(3/2)) ] dx from -a to +a

Evaluating this integral results in:

• E = (Q / (4πε₀a²))

Thus, the electric field at point A(0, a) is directed upwards along the y-axis with a magnitude of:

• **E = (Q / (4πε₀a²))**

Electric Potential at Point B(0, 0)

Now, let’s find the electric potential at point B(0, 0). The electric potential V due to a point charge is given by:

• V = (1 / (4πε₀)) * (Q / r)

For a continuous charge distribution, we can integrate the contributions from each charge element dq:

• V = ∫(dV) = ∫[ (1 / (4πε₀)) * (dq / r) ]

Distance from Charge Element to Point B

For a charge element located at position x, the distance to point B is simply:

• r = √(x² + 0²) = |x|

Substituting dq into the integral gives:

• V = (1 / (4πε₀)) * ∫[ (λ dx) / |x| ] from -a to +a

Substituting λ again results in:

• V = (Q / (8πε₀)) * ∫[ (1 / |x|) ] dx from -a to +a

Evaluating the Integral

This integral diverges at x = 0, but we can evaluate it as follows:

• V = (Q / (8πε₀)) * [ln(a) - ln(-a)] = (Q / (4πε₀)) * ln(2a)

Thus, the electric potential at point B(0, 0) is:

• **V = (Q / (4πε₀)) * ln(2a)**

Summary

In summary, we found that the electric field at point A(0, a) is:

• **E = (Q / (4πε₀a²))** (upwards along the y-axis)

And the electric potential at point B(0, 0) is:

• **V = (Q / (4πε₀)) * ln(2a)**

This approach illustrates how to apply the principles of electrostatics to find both the electric field and potential due to a continuous charge distribution. If you have any further questions or need clarification on any part, feel free to ask!