How to use the gradient to find Electric field

[howell]

1. A rod carrying a uniform charge distribution is bent into a semi circle with the center on the orgin and a radius R. Calcualte the Electric field at the center of the semi circle using the electric potential expression found in part a



2. E = -(gradient)V



3. The electric potential at the center is V = kQ/r or k (pi) (lambda) where lambda is the charge density. The correct expression for the electric field at the center of the circle is 2k(lambda)/r. However, I'm finding that simply taking the gradient of the electric potential function gives kQ/r^2 or k(lambda)(pi)/r, which is not correct. I'm afraid I'm making a fundamental mistake with the definition of the gradient and how to account for the changing electric field direction at the center of the semi circle.

 

[rl.bhat]

Here you have taken the gradient along the radius.
First you have to finds out delta Ex( due to a small element dl) = -dV/dx and dEy = - dv/dy where dx = dr*cos (theta) and dy = dr*sin(theta). Take summation over the whole semicircular rod to find the field.

 

[nlightner]

I would first like to applaud your efforts in attempting to use the gradient to find the electric field at the center of the semi circle. However, I can see where you may have made a mistake in your calculations.

The correct expression for the electric field at the center of the semi circle is indeed 2k(lambda)/r. This can be derived by considering the electric field as the negative gradient of the electric potential, as shown in your second equation.

However, the mistake may lie in your understanding of the gradient. The gradient is a vector operator that takes into account both the magnitude and direction of change in a function. In this case, the electric potential function changes in both magnitude and direction as you move along the semi circle. Therefore, the gradient at the center of the semi circle should be taken as a vector, rather than a scalar.

To find the correct expression for the electric field at the center of the semi circle, you should take the gradient of the electric potential function in terms of both x and y coordinates, and then add the two vectors together. This will give you the correct magnitude and direction of the electric field at the center of the semi circle.

In summary, to use the gradient to find the electric field at the center of the semi circle, you should consider the gradient as a vector operator and take the gradient in terms of both x and y coordinates. I hope this explanation helps you in your calculations. Keep up the good work!