Derive an expression between electric field and potential
Answers
Derive relation between electric field and electric potential
Deriving electric field from potential. ... In vector calculus notation, the electric field is given by the negative of the gradient of the electric potential, E = −grad V. This expression specifies how the electric field is calculated at a given point. Since the field is a vector, it has both a direction and magnitude.
Explanation:
heymate!! welcome to my answer....
To place a charge in the vicinity of an electric field, you should do work against the electrostatic force on the charge. This work done to bring a charge q to an electric field of some other charge configuration from infinity to a distance r, in the field is what we call the potential at the point r.
To do a work to move a charge q from a potential V to a small infinitesimal distance where the potential is V+dV, work has to be done against the Electric field.
The force acting on the charge q is
F=qE
So work done to move the charge through a potential difference of dV is:
dW=−F.dr
the negative sign implies work has to be done against the electrostatic force.
This work is the charge times potential difference between the points a and b (separated by a distance dr)
dW=−F.dr=−q(E.dr)=qdV
or
−E.dr=dV
or
E=−dVdr
dVdr is called the gradient of the scalar potential V.
Hence electric field is the negative gradient of the scalar potential. The negative sign came as a result because the potential difference is the work done per unit charge against the electrostatic force to move a charge from a to b.
However, this equation is valid only for static electrostatic fields.
The error in your math was that to calculate the work done, the displacement should be against the force. So you must put a negative sign as I did:
dW=−F.dr
hope its clr to u..