An electric field is produced by the very long, uniformly charged rod drawn above. If the strength of the electric field is E1 a
1 answer:
Answer: hello the complete question is attached below
answer :
r2 = 4r1
Explanation:
Electric field strength = F / q
we will assume the rod has an infinite length
For an infinitely charged rod
E ∝ 1/ r
considering two electric fields E1 and E2 at two different locations as described in the question
E1/E2 = r1/r2 ----- ( 2 )
<u>Calculate for r2 when E2 = E1/4 </u>
back to equation 2
E1 / (E1/4) = r1 / r2
∴ r2 = 4r1
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1) Magnitude of the force:
The magnetic field generated by a current-carrying wire is
where
is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated
Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
,
which is the distance at which the other wire is located:
Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)
Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive
A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
The magnetic field generated by a current-carrying wire is
where
I is the current in the wire
r is the distance at which the field is calculated
Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
which is the distance at which the other wire is located:
Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)
Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive
A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.