Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;
c) r>b E=ρ b (b-a)/r*εo
Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.
As we know,
∫E.dr= Qinside/εo
For r<a --->Qinside=0 then E=0
for a<r<b er have
E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L
E*2π*r*L =ρ*2*π*r* (r-a)*L/εo
E=ρ*(r-a)/εo
Finally for r>b
E*2π*r*L =ρ*2*π*b* (b-a)*L/εo
E=ρ*b* (b-a)*/r*εo
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Energy that transfers through the medium
The starter motor's potential difference across the headlight bulbs is 38.45V, requiring an additional 39 a from the battery. Voltage, also known as potential difference.
It is sometimes described as the amount of work needed to move a test charge between two sites, expressed as a unit of charge. Volt is the potential difference's SI unit (V). We only take into account the charge between the locations P and Q when current moves between them in an electric circuit. Electric potential difference between two sites is referred to as voltage, also known as electric pressure, electric tension, or (electric) potential difference. an electric field that is static.
Vh = I*Rn
Vh = 39/5.476*5.40v
Vh = 38.45v
Learn more about voltage here
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