Question

What is the maximum possible potential (relative to infinity) of the metal sphere of 10-cm radius? What is the maximum possible potential (relative to infinity) of the metal sphere of only 1-mm radius? These results hint at the reason why a highly charged piece of metal (with uniform potential throughout) tends to spark at places where the radius of curvature is small or at places where there are sharp points. Remember that breakdown electric strength for air is roughly\[{\bf{3 \times 1}}{{\bf{0}}^{\bf{6}}}\;\frac{{\bf{V}}}{{\bf{m}}}\].

Short answer

The maximum possible potential of metal spheres for radius \(10\;{\rm{cm}}\) and \(1\;{\rm{mm}}\) is \(3 \times {10^5}\;{\rm{V}}\) and \(3 \times {10^3}\;{\rm{V}}\).

Step-by-step solution

Write the given data

The breakdown electric strength of air is\(E = 3 \times {10^6}\;\frac{{\rm{V}}}{{\rm{m}}}\).

The radius of metal sphere is\(r = 10\;{\rm{cm}}\).

The radius of metal sphere is \(r = 1\;{\rm{mm}}\).

Conceptual Explanation

The breakdown strength of any medium is the effect due to which medium can hold maximum potential. The charge from medium starts leaking if potential become above the breakdown strength.

Determine the maximum possible potential of sphere

The maximum possible potential of the sphere is given as:

\(V = E \cdot r\)

Substitute\(r = 10\;{\rm{cm}}\)in the above equation.

\(\begin{array}{l}V = \left( {3 \times {{10}^6}\;\frac{{\rm{V}}}{{\rm{m}}}} \right)\left( {10\;{\rm{cm}}} \right)\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\\V = 3 \times {10^5}\;{\rm{V}}\end{array}\)

Substitute\(r = 1\;{\rm{mm}}\)in the above equation.

\(\begin{array}{l}V = \left( {3 \times {{10}^6}\;\frac{{\rm{V}}}{{\rm{m}}}} \right)\left( {1\;{\rm{mm}}} \right)\left( {\frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right)\\V = 3 \times {10^3}\;{\rm{V}}\end{array}\)