$$
begin{align*}
F_e= q_{T} cdot E
end{align*}
$$

where $q_{T}$ is charge of test charge, $F_e$ is force with which electric field of magnitude $E$ acts on test charge. We can express magnitude of the electric field $E$ from the equation above:

$$
begin{align*}
E= dfrac{F_e}{q_{T}}
end{align*}
$$

By measuring the force $F_e$ with which electric field $E$ acts on test charge $q_{T}$, we can determine magnitude of the present electric field.
To add up to the complexity, we can also determine the direction of the electric field lines so that electric field $E$ is completely described.
We can do this by determining the direction of force with which electric field acts on our test charge.

$$
F_e = q E
$$

$$
boxed{ text{ Direction of the electric field $E$ acting on test charge $q$ is eastward } }
$$

To find the magnitude of this electric field, we’ll use express electric field $E$ from the equation above:

$$
begin{align*}
F_e = q E \
tag{express $E$ from the equation above} \
E &= dfrac{F_e}{q} \
E &= dfrac{1.5 cdot 10^{-3} ~mathrm{N}}{2.4 cdot 10^{-8} ~mathrm{C}}
end{align*}
$$

$$
boxed{ E= 62500~mathrm{dfrac{N}{C}} }
$$

We found that the electric field is $E= 62500~mathrm{dfrac{N}{C}}$, eastward.

We know that magnitude of this electric field $E$ coming from charge $q$ at a distance $r$ from charge $q$ is given as:

$$
E = dfrac{kq}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.

As it can be seen, magnitude of the electric field that charge $q_{test}$ feels won’t be affected by magnitude of charge $q_{test}$, but only on charge $q$ and distance between charge $q_{test}$ and $q$.

Coulomb’s law states that the electric force $F$ between charges $q$ and $q_{test}$ at a distance $r$ from each other is given as:

$$
F = dfrac{k q q_{test}}{r^2}
$$

As it can be seen, magnitude of the electric force between these two charges will depend on magnitude of both of the charges $q$ and $q_{test}$ and distance $r$ between them. This means that force between these two charges does depend on $q_{test}$ whereas magnitude of the electric field that $q_{test}$ feels doesn’t depend on charge $q_{test}$.

We know that magnitude of this electric field $E_1$ coming from charge $q_1$ at a distance $r$ from it is given as:

$$
E_1 = dfrac{kq_1}{r^2}
$$

and that
magnitude of this electric field $E_2$ coming from charge $q_2$ at a distance $r$ from it is given as:

$$
E_2 = dfrac{kq_2}{r^2}
$$

where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.

As it can be seen, magnitude of the electric field that charge $q_{test}$ feels won’t be affected by magnitude of charge $q_{test}$, but only on charges $q_1$ and $q_2$ and distance between charge $q_{test}$ and the other two charges.

Net electric $E$ field felt by charge $q_{test}$ will be equal to:

$$
E = E_1 + E_2
$$

Keep in mind that if we want to measure a resulting electric field coming from charges $q_1$ and $q_2$, we need to use a test charge $q_{test}$ with such a low charge magnitude that it almost won’t affect the electric field coming from the two negative charges. If this test charge affects or distorts the electric field of any of the two negative charges by a lot, we don’t essentially measure the electric field produced by the two negative charges because this field will be so distorted and unmeasurable with this test charge. Electric field lines between two negative charges should look the same way as they look in figure 24-b and while positive charge will inevitably bring some distortion, this distortion must not be so great as shown in figure 21-2c.