I have a question about the definition of electric field. Texts define it as the force on on a positive test charge ($q_0$), divided by the charge on the positive test charge.

$$ E = k \frac{q_1q_0}{q_0r^2}$$

My question is that since we are dividing out the test charge, couldn't a test charge of negative value also be used to obtain an identical electric field?

I do see that in the equation $F = qE$, only a positive $q$ experiences a force in the same direction as the field. But it seems like the identical field could be obtained by using either positive or negative test charges.


2 Answers 2


You are correct: If $F(q_1, q_2)$ is the force $q_1$ experiences under $q_2$, and you define

$$E(q_2) \equiv \lim_{q_1 \to 0} \frac{F(q_1,q_2)}{q_1}$$

where $E(q_2)$ is the electric field due to a particle $q_2$, then it doesn't matter where $q_1$ is positive or negative, i.e., whether you take the limit from the left or the right.

  • $\begingroup$ As I said in answer to the same question asked a few weeks ago, electric field strength is defined in such a way (that is including the test charge in the denominator) that it is independent of either the magnitude or the direction of the test charge itself, but depends only on the environment in which the charge is placed. $\endgroup$ Sep 21, 2017 at 21:45

Update as a result of a comment.

There are two ways of answering the question.

The first is to say that the magnitude and direction of the force on a unit test charge is the electric field.
If this was the definition then the direction of the force would differ as to whether the test charge was negative or positive,
$\vec E= \dfrac {\vec F}{q\,(\text{no sign assigned)}}$
A positive test charge would have a force on it in one direction and a negative test charge would have a force on it in the opposite direction.

However that definition is not the one that is used.
When using a unit test charge there must be included in the definition of the electric field the sign of the unit test charge.
$\vec E= \dfrac {\vec F}{\pm q\,(\text{sign assigned)}}$

As you have pointed out when this definition is used it matters not what the sign of the test charge is, the direction of the electric field is always the same.
If the direction of the force on a positive test charge is positive then the electric field is in the positive direction.
The force on a negative charge at the same position would be in the negative direction but the direction of the electric field would be positive as there would be a negative sign in both the denominator and numerator of the equation which defines electric field.

  • $\begingroup$ Can you explain why? If q1 is positive and the test charge is positive, the force felt by the test charge and the electric field are both in the positive radial direction. $\endgroup$
    – lamplamp
    Sep 21, 2017 at 18:15
  • $\begingroup$ BUT if q1 is positive and the test charge is negative, then the force felt by the test charge is in the negative direction. What is confusing me is that it seems to me that the electric field, (defined in the alternate way I'm hypothetically proposing by dividing the force felt by negative test charge by the value of the negative charge), is still in the positive direction since the negative signs of the test charge cancel out in numerator and denominator. $\endgroup$
    – lamplamp
    Sep 21, 2017 at 18:19

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