I cross referenced some website yahoo answers, wikipedia & some other websites. They were providing different answers. I know that electric field intensity is a vector quantity. But what abt electric field. Are both electric field and electric field potentail same or different. 
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asked Jun 28, 2015 at 11:31
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2$\begingroup$ You should include the definitions of "vector quantity" and "scalar quantity" you are using, because they seem to be backwards from what I usually would say... $\endgroup$– MartinCommented Jun 28, 2015 at 12:35
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3 Answers
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The electric field, as already pointed out by @Mew, is defined dividing the force by the charge you are using to measure the force upon $$ \textbf{E} = \frac{\textbf{F}}{q} $$ and as such, whatever the force is (Coulomb or whatsoever else) it is a vector. The potential of a vector field $\textbf{v}$ is, by definition in mathematics, a twice differentiable function $f(x,y,z)$ such that, at any point within the domains of definitions $$ \textbf{v}(x,y,z)= \textrm{grad}\,f(x,y,z) $$ namely $df(x,y,z) = v_xdx + v_ydy + v_z dz$. Once you have understood what the definitions are, you can go ahead interpreting the two as force per unit charge and change in the force per unit charge; however, from the above definitions it is clear what their nature is.
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The electric field is a vector quantity, representing the electric force per unit charge acting on a test particle at a particular position in space. Since force is a vector, the electric field too is a vector quantity.
The electric potential however is not a vector. The electric potential is the amount of electric potential energy that a unitary point electric charge would have if located at any point in space, and energy is a scalar quantity.
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$\begingroup$ A test charge is being used to determine the potential of the charge . It should be vector quantity.. $\endgroup$ Commented Jun 28, 2015 at 11:37
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$\begingroup$ @AUmarMukthar, remind me again how the definition of a vector is that a test charge is used? I thought a vector was something with a magnitude and a direction. $\endgroup$– KenshinCommented Jun 28, 2015 at 11:54
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1$\begingroup$ @AUmarMukthar A test charge is not being used to determine the potential. A test charge is being used to determine the field, whose associate differential form is the potential. $\endgroup$– gentedCommented Jun 28, 2015 at 13:57
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$\begingroup$ It is electric field intensity.. typo error.. $\endgroup$ Commented Jun 28, 2015 at 14:26
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$\begingroup$ @AUmarMukthar, well that one is a vector. Also electric field is not the same as electric field potential. They describe two different quantities and also the latter is a scalar while the former is a vector. $\endgroup$– KenshinCommented Jun 29, 2015 at 7:51
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"Scalar", "vector", "tensor" and so on describe properties of an object under some set of transformations and depend on this set. For example, electric field is a vector under transformations from the group of 3-dimensional rotations and a part of a tensor under Lorentz transformations. In the same way, the electric potential is a scalar under rotations and a part of a vector under Lorentz transformations.
answered Jun 28, 2015 at 15:55