If we assume potential at infinity to be zero and earth to be spherical then the potential at the surface of the earth is given by $\frac{kq}{r}$,,where $k$ is a constant  ,$q$ is the charge on earth and $r$ its radius.As $q$ is extremely small and $r$ very large, so the potential at earth's surface is almost zero.So for all practical purposes we assume its potential to be zero.The potential at infinity is assumed to be absolutely zero whereas that on earth's surface almost zero.

If we assume potential at infinity to be zero and Earth to be spherical then the potential at the surface of the earth is given by ${kq}/{r}$, where $k$ is a constant,  $q$ is the charge on earth and $r$ its radius. As $q$ is extremely small and $r$ very large, the potential at Earth's surface is almost zero. So for all practical purposes we assume its potential to be zero. The potential at infinity is assumed to be absolutely zero whereas that on Earth's surface almost zero.

If we assume potential at infinity to be zero and earth to be spherical then the potential at the surface of the earth is given by kq/r,,where k is a constant ,q is the charge on earth and r its radius.As q is extremely small and r very large, so the potential at earth's surface is almost zero.So for all practical purposes we assume its potential to be zero.The potential at infinity is assumed to be absolutely zero whereas that on earth's surface almost zero. Sanjay Bisht

If we assume potential at infinity to be zero and earth to be spherical then the potential at the surface of the earth is given by $\frac{kq}{r}$,,where $k$ is a constant ,$q$ is the charge on earth and $r$ its radius.As $q$ is extremely small and $r$ very large, so the potential at earth's surface is almost zero.So for all practical purposes we assume its potential to be zero.The potential at infinity is assumed to be absolutely zero whereas that on earth's surface almost zero.