The equipotential surface always perpendicular to the electric field because
1.work done on surface is zero four point charge
2. Work done never equal to zero
3. Electric field is conservative
4. None of these
The equipotential surface always perpendicular to the electric field because
1.work done on surface is zero four point charge
2. Work done never equal to zero
3. Electric field is conservative
4. None of these
In an electrostatic field, the force acting on a charge inside the field is tangent to the lines of force. Suppose you move this charge in an arbitrary direction: the work to be done to perform this movement equals the scalar product of the electrical force on the charge and the displacement vector.
Now, being the field potential, this work is also equal to the difference of potential between the initial and final position of the charge.
As the scalar product of two vectors is zero when the vectors are perpendicular, then if you move the aforementioned charge in a direction perpendicular to the lines of force, then you work is zero and the potential remains the same.
As equipotential surfaces are by definition surfaces where the potential remains constant, then they must be perpendicular to the lines of force exactly for what I explained before.
So option (1) is the answer.
In an electrostatic field, the force acting on a charge inside the field is tangent to the lines of force. Suppose you move this charge in an arbitrary direction: the work to be done to perform this movement equals the scalar product of the electrical force on the charge and the displacement vector.
Now, being the field potential, this work is also equal to the difference of potential between the initial and final position of the charge.
As the scalar product of two vectors is zero when the vectors are perpendicular, then if you move the aforementioned charge in a direction perpendicular to the lines of force, then you work is zero and the potential remains the same.
As equipotential surfaces are by definition surfaces where the potential remains constant, then they must be perpendicular to the lines of force exactly for what I explained before.
So option (1) is the answer.
