It is said that a charged particle has it's own electric field even in the absence of other charged particles, but how could one verify this statement? As to verify this one must place a charge particle in it's surrounding.
My mind is telling me " A charge cannot produce Electric field on it's own, it needs another charge to produce an electric field ". I know this statement is wrong but please provide me a contradiction along with a proof.
a charged particle has it's own electric field even in the absence of other charged particles, but how could one verify this statement?
A large part of the mass of a charged particle is in the energy of the electric field. If the field disappears then the mass changes dramatically. So in principle you could use collisions with neutrinos to test if the mass has suddenly changed.
I propose an answer complementary to the one by @Dale.
From the conceptual point of view, the electric field is defined as the limit for vanishing charge of the electric force on a test charge $q$.
$$ {\bf E} = \lim_{q\rightarrow 0} \frac{{\bf F}}{q}. \tag{1} $$
This definition is necessary because a finite test charge acts on the sources of the force field, modifying them. Only when the test charge is so small that the induced modification can be neglected can we write the force as the product of the test charge by a vector whose sources are independent of the test charge.
Therefore, the most direct verification of the existence of the electric field corresponds to checking the physical existence of the limit in formula ($1$) as a vector independent of $q$.
