a) If a proton moved from a location with a 5.0 V potential to a location with 7.5 V potential, would its potential energy incre
1 answer:
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Explanation:
Given that,
Length of gold wire, l = 4 m
Voltage of battery, V = 1.5 V
Current, I = 4 mA
The resistivity of gold,
Resistance in terms of resistivity is given by :
Also, V = IR
So,
A is area of wire,
, r is radius, r = d/2 (diameter=d)
Out of four option, near option is (C) 17 μm.
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² =
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and
for
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
