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  5. Electric field

5.4 Electric field  (Page 3/7)

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The direction of the field

[link] enables us to determine the magnitude of the electric field, but we need the direction also. We use the convention that the direction of any electric field vector is the same as the direction of the electric force vector that the field would apply to a positive test charge placed in that field. Such a charge would be repelled by positive source charges (the force on it would point away from the positive source charge) but attracted to negative charges (the force points toward the negative source).

Direction of the electric field

By convention, all electric fields E → point away from positive source charges and point toward negative source charges.

Add charges to the Electric Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

The E -field of an atom

In an ionized helium atom, the most probable distance between the nucleus and the electron is r = 26.5 × 10 −12 m . What is the electric field due to the nucleus at the location of the electron?

Strategy

Note that although the electron is mentioned, it is not used in any calculation. The problem asks for an electric field, not a force; hence, there is only one charge involved, and the problem specifically asks for the field due to the nucleus. Thus, the electron is a red herring; only its distance matters. Also, since the distance between the two protons in the nucleus is much, much smaller than the distance of the electron from the nucleus, we can treat the two protons as a single charge +2 e ( [link] ).

A positive charge of plus 2 e is shown at the center of a sphere of radius r. An electron is depicted as a particle on the sphere. The vector r is shown as a vector with its tail at the center and its head at the location of the electron. The electric field at the location of the electron is shown as a vector E with its tail at the electron and pointing directly away from the center.
A schematic representation of a helium atom. Again, helium physically looks nothing like this, but this sort of diagram is helpful for calculating the electric field of the nucleus.

Solution

The electric field is calculated by

E → = 1 4 π ε 0 ∑ i = 1 N q i r i 2 r ^ i .

Since there is only one source charge (the nucleus), this expression simplifies to

E → = 1 4 π ε 0 q r 2 r ^ .

Here q = 2 e = 2 ( 1.6 × 10 −19 C ) (since there are two protons) and r is given; substituting gives

E → = 1 4 π ( 8.85 × 10 −12 C 2 N · m 2 ) 2 ( 1.6 × 10 −19 C ) ( 26.5 × 10 −12 m ) 2 r ^ = 4.1 × 10 12 N C r ^ .

The direction of E → is radially away from the nucleus in all directions. Why? Because a positive test charge placed in this field would accelerate radially away from the nucleus (since it is also positively charged), and again, the convention is that the direction of the electric field vector is defined in terms of the direction of the force it would apply to positive test charges.

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The E -field above two equal charges

(a) Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges + q that are a distance d apart ( [link] ). Check that your result is consistent with what you’d expect when z ≫ d .

(b) The same as part (a), only this time make the right-hand charge − q instead of + q .

Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta.
Finding the field of two identical source charges at the point P . Due to the symmetry, the net field at P is entirely vertical. (Notice that this is not true away from the midline between the charges.)

Strategy

We add the two fields as vectors, per [link] . Notice that the system (and therefore the field) is symmetrical about the vertical axis; as a result, the horizontal components of the field vectors cancel. This simplifies the math. Also, we take care to express our final answer in terms of only quantities that are given in the original statement of the problem: q , z , d , and constants ( π , ε 0 ) .

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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