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Consider a square with side a. We put the charges q, -q, 2q, -2q in the corners of this square. Check the relation E=-∇ϕ at the center of the square

Question

Consider a square with side a.

We put the charges q, -q, 2q, -2q in the corners of this square. Check the relation E=ϕ E = -\nabla \phi at the center of the square.

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Solution

To check the relation E=-∇ϕ at the center of the square, we first need to understand the terms involved:

  1. E is the electric field, which is a vector field that shows the force a positive test charge would experience at any given point in space.
  2. ∇ϕ is the gradient of the electric potential ϕ, which is a scalar field that shows the potential energy per unit charge.

The negative sign in the relation E=-∇ϕ indicates that the electric field points in the direction of decreasing electric potential.

Now, let's proceed with the steps:

Step 1: Calculate the electric potential at the center of the square

The electric potential ϕ at a point due to a point charge q is given by ϕ = kq/r, where k is Coulomb's constant and r is the distance from the charge to the point. At the center of the square, the distance to each corner is √2a/2 = a/√2. Therefore, the total electric potential at the center is the sum of the potentials due to each charge:

ϕ = kq/(a/√2) - kq/(a/√2) + 2kq/(a/√2) - 2kq/(a/√2) = 0

Step 2: Calculate the electric field at the center of the square

The electric field E at a point due to a point charge q is given by E = kq/r², where r is the distance from the charge to the point. At the center of the square, the distance to each corner is a/√2. Therefore, the total electric field at the center is the vector sum of the fields due to each charge:

E = kq/(a/√2)² - kq/(a/√2)² + 2kq/(a/√2)² - 2kq/(a/√2)² = 0

Step 3: Check the relation E=-∇ϕ

Since both the electric field E and the gradient of the electric potential ∇ϕ are zero at the center of the square, the relation E=-∇ϕ holds at this point.

This problem has been solved

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