The angle between the electric lines of force and the equipotential surface is:

Concept: 

We have considered an equipotential surface (a surface on which potential is constant everywhere).

\(\overline n\) = normal component through the surface 

\(\vec{E}\) = electric field through a surface

θ = angle between electric field and the surface of the sheet

here we know that  potential difference = - \(\int E^{\vec{}}. dr^{\vec{}} \)  ..........(1)

Calculations: 

Here as we know that this is an equipotential surface then dV = 0

or, dV = -\(\int E^{\vec{}}. dr^{\vec{}} \) = 0 

from here we get \(E_{per}.dr = 0 \)

which says that the scalar product of the normal component of the electric field with distance is zero.

so that we can consider the normal component of the Electric field as = E cosθ 

now, E cosθ . dr = 0

i.e. E and dr both are non-zero values

∴ cosθ = 0 ⇒ θ = 90o

Now θ = 90^o means that the angle between electric lines of forces or electric field and the equipotential surface is 90⁰. 

So option 4) is correct.