26: PhET Tutorial: Charges and Electric Fields

Part A: 
The electric field produced by the positive charge

a) is directed radially toward the charge at all locations near the charge.
b) wraps circularly around the positive charge.
c) is directed radially away from the charge at all locations near the charge.

SOLUTION:
This means that another positive charge, if placed near the original charge, would experience a force directed radially away from the original charge.

Part B:
For these four locations, the magnitude of the electric field is

a) greatest to the left of the charge.
b) the same.
c) greatest below the charge.
d) greatest to the right of the charge.
e) greatest above the charge.

SOLUTION:
This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge. Mathematically, we say the electric field is spherically symmetric.

Part C:
The magnitude of the electric field 1 m away from the positive charge is 
a) one-half
b) equal to
c) two times
d) four times
e) one-quarter
the magnitude of the electric field 2 m away.

SOLUTION:
The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared (E∝1/r², where r is the distance from the charge). You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulomb'€™s law, which states that the magnitude of the force between two charged particles is F=kQ₁Q₂/r².

Part D:
If the field strength is E = 9 V/m a distance of 1 m from the charge, what is the field strength E a distance of 3 m from the charge?

SOLUTION:
E∝1/r², so E₂ = 9/3² = 1 V/m

Part E:
Which statement best describes the differences in the electric field due to a negative charge as compared to a positive charge?

a) The electric field changes direction (now points radially inward), but the electric field strength does not change.
b) Nothing changes; the electric field remains directed radially outward, and the electric field strength doesn'™t change.
c) The electric field changes direction (now points radially inward), and the magnitude of the electric field decreases at all locations.

SOLUTION:
The electric field is now directed toward the negative charge, but the field strength doesn'€™t change. The electric field of a point charge is given by E⃗ =(kQ/r²)r^. Because of the sign of the charge, the field produced by a negative charge is directed opposite to that of a positive charge but the magnitude of the field is the same.

Part F:
Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)?

a) The electric field is nonzero everywhere on the screen.
b) The electric field is roughly zero near the midpoint of the two charges.
c) The electric field is zero at any location along a vertical line going through the point directly between the two charges.

SOLUTION: 
Directly between the two charges, the electric fields produced by each charge are equal in magnitude and point in opposite directions, so the two vectors add up to zero.

Part G:
Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 V/m. What is the magnitude of the total electric field due to both charges at this location?

a) 36 V/m
b) 25 V/m
c) zero

SOLUTION:
Notice that this number is less than twice the magnitude of the field due to each charge. This occurs because the horizontal components of the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together.

Part H:
The electric field at the midpoint is

a) directed to the left
b) zero
c) directed to the right

SOLUTION:
The electric field due to the positive charge is directed to the right, as is the electric field due to the negative charge. So the net electric field, which is the sum of these two fields, is also to the right.

Part I:
Measure the strength of the electric field 0.5 m directly above the midpoint as well as 1 m directly above. Does the strength of the electric field decrease as 1 over distance squared (1/r²)?

a) No, it decreases less quickly with distance
b) No, it decreases more quickly with distance
c) Yes, it does

SOLUTION:
In fact, it turns out that the strength of the electric field decreases roughly as 1/r³! So the field 1 m above the midpoint is roughly eight times weaker than at 0.5 m above the midpoint. The important lesson here is that, in general, a distribution of charges produces an electric field that is very different from that of a single charge.

Part J:
Measure the strength of the electric field 1 m directly above the middle as well as 2 m directly above. Does the strength of the electric field decrease as 1 over distance squared (1/r²)?

a) No, it decreases less quickly with distance
b) No, it decreases more quickly with distance
c) Yes, it does

SOLUTION:
In fact, it turns out that the strength of the electric field decreases roughly as 1/r. So the field 1 m above the midpoint is roughly half the strength at 0.5 m. This is another example showing that a distribution of charges produces an electric field that is very different from that of a single charge.