If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
1 answer:
Explanation:
The electric field of an isolated charged parallel-plate capacitor is given by :
........(1)
Where
q is the electric charge
A is the area of cross section of parallel plate
It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.
So, the correct option is (E) i.e. "none of the above".
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Answer:
The magnitude of magnetic field at given point = ×
T
Explanation:
Given :
Current passing through both wires = 5.0 A
Separation between both wires = 8.0 cm
We have to find magnetic field at a point which is 5 cm from any of wires.
From biot savert law,
We know the magnetic field due to long parallel wires.
⇒
Where magnetic field due to long wires,
,
perpendicular distance from wire to given point
From any one wire 5 cm,
3 cm
so we write,
∴
Therefore, the magnitude of magnetic field at given point =