A non conducting semicircular disc of radius b with a cut out semicircular disc of radius a has a uniform charge density sigma. What''s the ratio of Electric field to electric potential at the centre of disc?

To find the ratio of the electric field to the electric potential at the center of a non-conducting semicircular disc with a cut-out semicircular disc, we need to analyze the contributions of both the full semicircular disc and the cut-out semicircular disc. Let's break this down step by step.

Understanding the Setup

We have a non-conducting semicircular disc of radius \( b \) with a smaller semicircular cut-out of radius \( a \). The entire structure has a uniform surface charge density \( \sigma \). The goal is to determine the electric field \( E \) and the electric potential \( V \) at the center of the disc, and then find their ratio.

Electric Field Calculation

The electric field at the center of a uniformly charged semicircular disc can be derived from the contributions of each infinitesimal charge element. For a semicircular disc, the electric field points along the axis of symmetry (the vertical line through the center).

• The electric field due to the full semicircular disc of radius \( b \) at the center is given by:

  E_b = \frac{\sigma}{2\epsilon_0}.

• The electric field due to the cut-out semicircular disc of radius \( a \) at the center is:

  E_a = \frac{\sigma}{2\epsilon_0}.

Since the cut-out disc has a negative charge density (it removes charge from the full disc), the total electric field at the center becomes: E = E_b - E_a = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0.

Electric Potential Calculation

The electric potential at a point due to a charged body is the work done in bringing a unit positive charge from infinity to that point. For a semicircular disc, the potential at the center can be calculated using the formula:

• The potential due to the full semicircular disc of radius \( b \) is:

  V_b = \frac{\sigma b}{2\epsilon_0}.

• The potential due to the cut-out semicircular disc of radius \( a \) is:

  V_a = \frac{\sigma a}{2\epsilon_0}.

Thus, the total electric potential at the center is: V = V_b - V_a = \frac{\sigma b}{2\epsilon_0} - \frac{\sigma a}{2\epsilon_0} = \frac{\sigma (b - a)}{2\epsilon_0}.

Finding the Ratio

Now that we have both the electric field \( E \) and the electric potential \( V \), we can find the ratio of electric field to electric potential at the center:

Ratio = \frac{E}{V} = \frac{0}{\frac{\sigma (b - a)}{2\epsilon_0}} = 0.

Final Thoughts

The ratio of the electric field to the electric potential at the center of the semicircular disc with a cut-out is zero. This result stems from the fact that while the potential remains finite due to the presence of charge, the electric field cancels out completely due to the symmetry of the charge distribution. This illustrates an interesting aspect of electrostatics where the potential can exist even when the electric field is zero.