How much force a proton will experience in an electric field of 5.1 x 10 ^(11) N/C ?
Answers
Explanation:
A field is a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding the earth (and all other masses) represents the gravitational force that would be experienced if another mass were placed at a given point within the field.
In the same way, the Coulomb force field surrounding any charge extends throughout space. Using Coulomb’s law,
F
=
k
∣
q
1
q
2
∣
r
2
F=k∣q1q2∣r2, its magnitude is given by the equation
F
=
k
∣
q
Q
∣
r
2
F=k∣qQ∣r2, for a point charge (a particle having a charge Q) acting on a test charge q at a distance r (see Figure 1). Both the magnitude and direction of the Coulomb force field depend on Q and the test charge q.
In part a, two charges Q and q one are placed at a distance r. The force vector F one on charge q one is shown by an arrow pointing toward right away from Q. In part b, two charges Q and q two are placed at a distance r. The force vector F two on charge q two is shown by an arrow pointing toward left toward Q.
Figure 1. The Coulomb force field due to a positive charge Q is shown acting on two different charges. Both charges are the same distance from Q. (a) Since q1 is positive, the force F1 acting on it is repulsive. (b) The charge q2 is negative and greater in magnitude than q1, and so the force F2 acting on it is attractive and stronger than F1. The Coulomb force field is thus not unique at any point in space, because it depends on the test charges q1 and q2 as well as the charge Q.
To simplify things, we would prefer to have a field that depends only on Q and not on the test charge q. The electric field is defined in such a manner that it represents only the charge creating it and is unique at every point in space. Specifically, the electric field E is defined to be the ratio of the Coulomb force to the test charge:
E
=
E
q
E=Eq,
where F is the electrostatic force (or Coulomb force) exerted on a positive test charge q. It is understood that E is in the same direction as F. It is also assumed that q is so small that it does not alter the charge distribution creating the electric field. The units of electric field are newtons per coulomb (N/C). If the electric field is known, then the electrostatic force on any charge q is simply obtained by multiplying charge times electric field, or F = qE. Consider the electric field due to a point charge Q. According to Coulomb’s law, the force it exerts on a test charge q is
F
=
k
∣
q
Q
∣
r
2
F=k∣qQ∣r2. Thus the magnitude of the electric field, E, for a point charge is
E
=
∣
∣
F
q
∣
∣
=
k
∣
∣
q
Q
q
r
2
∣
∣
=
k
|
Q
|
r
2
E=|Fq|=k|qQqr2|=k|Q|r2.
Since the test charge cancels, we see that
E
=
k
|
Q
|
r
2
E=k|Q|r2.
The electric field is thus seen to depend only on the charge Q and the distance r; it is completely independent of the test charge q.
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