Why can't an electric field line suddenly break? My book has the following question in it as an exercise:

An electrostatic field line is a continuous curve i.e. a field line cannot have sudden breaks. Why not?

I cannot seem to be able to reason for or against this statement.

*

*Can somebody please tell me why they can't suddenly break and if that were to happen then what will be its outcome?

 A: Electric field lines have no physical existence: they are useful concepts to understand vector fields, but they do not carry an independent physical ontology of their own.
Given a vector field $\mathbf E(\mathbf r)$, we define the field lines as the solutions of the differential equation
$$
\frac{\mathrm d\boldsymbol \gamma}{\mathrm ds} = \mathbf E(\boldsymbol \gamma(s)),
\tag 1
$$
i.e., as the continuous curves whose derivative is given by the electric field at the curve. (This is up to re-parametrizations of the curve, which turn the equation above into a proportionality, but which don't affect the geometric locus of the field line, which is ultimately the only thing we care about.)
The definition $(1)$ means that field lines can have kinks where their derivative is discontinuous if they meet places where the electric field is discontinuous, such as at a surface charge.
However, field lines cannot break because we define them as continuous objects: basically, putting a pen at a starting point and then following the arrows of the vector field without lifting the pen.
A: Charges are "sources" and "sinks" of electric field lines. If a field line was "broken" it would indicate the presence of a charge. If this were not the case then Gauss's law would not hold. Gauss's law guarantees continuous field lines except at the location of charges.
A: My answer is complementary to that given by @BioPhysicist, who looked at the problem from the point of view of the integral form of the Gauss law. Its differential form says:
$$
\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0},
$$
which means the electric field has finite derivative everywhere except singular points/surfaces, which means that the electric field itself continuous, i.e. the electric field lines cannot interrupt. However the derivative of electric field can experience jumps, i.e. the electric field lines can be broken, but this requires singular charge distributions (point charges, charged wires of zero thickness or infinitely thin charged planes) or sharp material boundaries.
A: Electric Fields, E are continuous gradients measured in volts/m [V/m] since this is due to the measurement of a voltage at a distance via the air medium which acts a variable attenuator according to geometry from a plate , line or point source that varies continuously by gap respectively as $$ k/r, k/r^2, k/r^3 $$
For a point source, attenuation results are defined by a Friis Loss formula for a defined source and receiver of measurement.
Conclusion
In  order to have a discontinuity in E field , consider you need to have a discontinuity in space.
