Electric field strength(#E#) at a particular point in an Electric field(being generated by a charge #Q#), is defined as:
the Force(#F#) that acts on a unit positive change(#q#) placed at that point.
That is, #E=F/q=1/(4piepsilon)(Qq)/r^2xx1/q=1/(4piepsilon)Q/r^2#
The magnitude of the Electric field(#E#) can also be viewed as being the negative of the Potential gradient#((dV)/(dr))#
So #" , "E=-(dV)/(dr) color(red)" {This is the relation i think}"#
Because #NC^-1-=Vm^-1#
where #V# in #(dV)/(dr)# is the Electric potential at the point under consideration,
And Electric Potential(#V#) at a point in the E-field is the work done in bring a unit charge form infinity to that point
#=> V=int_oo^(r_1)1/(4piepsilon)(Qq)/r^2xx1/q-=int_oo^(r_1)Edr#
#=>V=-1/(4piepsilon)(Q)/(r_1)#
So, it is logical that if, the #V=-intE#, then the derivative #(dV)/(dr)=-E#
