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Yes it does have external electric field between its two terminals.
If there are two points with a potential difference V separated apart by a distance d (in a dielectric like air or vacuum) then
the electric field is given by E = V/d.
If the geometry is complicated, corrections and appropriate relations apply.
Many obvious things are difficult to locate as tutorials or detailed literature.
You can start your search from here
Electric Potential Difference
Once you agree the existence of potential between two points separated apart in a dielectric, the presence of electric field follows.
Hope this helps.
Yes, and to a first approximation, if the battery is 51 mm (0.051 m) long, as stated at AA battery - Wikipedia, the free encyclopedia
and the voltage is 1.5 V, then the strength of the field is 1.5/0.051 = 29.4 V/m. Electric field is a vector quantity (has magnitude and direction), and will point from the negative to positive terminal, as for a uniform field in the x direction, the electric field is -dV/dx. Note the minus sign.
These are all very rough approximations - especially for a battery due to its shape.
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