0
$\begingroup$

Can someone verify whether this is correct?

Given the Coulomb force, we can define the electric field in a point in space as $\bar E = \frac{\bar F}{q} $, where q is a positive test charge and $\bar F$ is the Coulomb force acting on q. We make the hypothesis that q does not make the other charges to move. Rewriting the equation above, when a point charge Q acts on q, we find that $\bar E = \frac{1}{4 \pi \epsilon_0} \frac{Q (\bar r_q - \bar r_Q)}{|\bar r_q - \bar r_Q|^3}$. Hence, we can interpret the electric field as the force, that comes from a point charge Q, acting on a charge of $1C$ (we can regognize this in the formula above). Is this correct?

From wikipedia:

The electric field, ${\displaystyle \mathbf {E} }$, at a given point is defined as the (vector) force, ${\displaystyle \mathbf {F} }$, that would be exerted on a stationary test particle of unit charge by electromagnetic forces (i.e. the Lorentz force). A particle of charge ${\displaystyle q}$ would be subject to a force ${\displaystyle \mathbf {F} =q\mathbf {E} }$.

Do they mean the magnitude of the charge of a proton with unit charge? Or 1 coulomb?

I'm very confused. Can someone clearly explain what electric field is if my thinking about it is wrong? Is there an intuitive way to think about it?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ They mean 1 Coulomb. The charge on a proton is many orders of magnitude smaller. $\endgroup$
    – Lewis Miller
    Commented Mar 21, 2017 at 15:46

1 Answer 1

Reset to default
2
$\begingroup$

Electric field at any point is the force that a unit positive charge (1unit positive charge= 1Coulomb) would feel when placed at that particular point. This is a good way to decide the direction of the electric field lines.

Basically we define electric field in terms of the force a particle experiences at that point because that's the only thing we can measure. Field lines are just there to help us visualise. Yet they are a very important concept in physics. They stress the importance of the finite time taken for propagation of electromagnetic waves in more advanced physics.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ But wouldn't the test charge change the field if you would place it somewhere? $\endgroup$
    – user147033
    Commented Mar 21, 2017 at 16:05
  • 1
    $\begingroup$ @Math_QED Very good thinking. Usually when we calculate the fields due to a source charge, we make certain assumptions. The first one of which is that the source charge is sufficiently 'large' in magnitude that its field lines are not distorted by the test charge (or alternatively you can say that the test charge is very small ~dq). Furthermore we assume that the source charge is held fixed at wherever it is by certain unspecified forces. $\endgroup$
    – Kunal Pawar
    Commented Mar 21, 2017 at 16:11
  • $\begingroup$ The second assumption usually follows from the first one (that the source charge is much much bigger in magnitude than the test charge). Think of anology. A boulder won't budge if you threw a pebble at it! Similarly bringing a very small test charge won't make the big source charge move. $\endgroup$
    – Kunal Pawar
    Commented Mar 21, 2017 at 16:14
  • 1
    $\begingroup$ Furthermore, @Math_QED, to add upon Kunal's explanation, remember that whatever field the test charge sets up, this does not influence that test charge itself - that only influences other charges. When looking at the effect on a test charge, its own field is not relevant. We only have to assume that this field doesn't change the surroundings (therefore the assumptions, which Kunal points out). $\endgroup$
    – Steeven
    Commented Mar 21, 2017 at 16:15
  • 1
    $\begingroup$ @Math_QED Yes and yes. It doesn't affect the big source charge whose field we are measuring (the field set up, as you say). Remember that a test charge is something we use to measure the field of a source charge. $\endgroup$
    – Kunal Pawar
    Commented Mar 21, 2017 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.