A positive test charge is released from rest in a uniform electric field. Which of the following best describes its initial motion?
A. It will move with a constant velocity along an equipotential line
B. It will move with constant acceleration along an equipotential line
C. It will move with constant acceleration along an electric field line
D. It will move with constant velocity along an electric field line

Hint: If an external electric field is applied and a test charge is kept inside the electric field, the force acting per unit charge describes the intensity of the electric field. This can give us the expression for force acting on the test charge.

Formula used:
The (intensity of) electric field is force per unit charge or:
$\vec{E} = \dfrac{\vec{F}}{q}$

Complete step by step answer:
To find the electric field intensity at a point we place a positive test charge q at that point and the coulomb force it experiences due to the field gives us the electric field intensity:
$\vec{E} = \dfrac{\vec{F}}{q}$
From this expression we can write:
$q \vec{E} = \vec{F}$
This clearly shows that a force is acting on the test charge and the direction of this force is same as that of the electric field.
Now if the charge is released, it will experience acceleration due to this force, which can be found easily as:
$\vec{F} = m \vec{a} = q \vec{E}$ .
So, we obtain:
$\vec{a} = \dfrac{q \vec{E}}{m}$
This expression really sorts everything out. It clearly shows that the acceleration that the test charge will experience will be in the same direction as the electric field.
Therefore the correct answer is option (C).

Additional information:
The fact that charged particles are accelerated by electric fields is used in designing various linear accelerators. A potential difference is applied across two plates and electrons emitted (thermally) are made to gain energy by accelerating in electric fields.

Note:
The direction of force has only two possibilities- along and opposite to the direction of the applied electric field. If there was a negative charge, we would have just replaced q with -q and found that the force of the negative charge would be opposite to the applied electric field direction. Therefore, a positive charge gets repelled by an electric field and negative charge gets attracted by the electric field.