What is Kinetic Energy and Potential energy❓Ty!!
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The electric field at arbitrary point outside the sphere is determined as the E = σr³/k.
<h3>Electric field determined from Gauss law</h3>The electric field of the surface is determined from Gauss law as shown below;
E ∫ds = Q/ε
E (4πr²) = Q/ε
E = Q/4πεr² . r
where;
- σ is dipole moment of atom of the metal
- k is dielectric constant
Thus, the electric field at arbitrary point outside the sphere is determined as the E = σr³/k.
The complete question:
A metal sphere of radius R carries a total charge Q. outside the sphere is a dielectric with polarization p(f) k/r^3er. Determine the electric field at arbitrary point outside the sphere.
Learn more about electric field here: brainly.com/question/14372859
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Answer: Heat current through the insulator=196W
Electric power= 196W
Explanation: Given: Kglass = 0.040W/m
Temperature of inside glassTi=175°C
Temperature of outside glass To= 35°C
Area=1.4m^2 , L= 4×10^-2
Heat current(H)= K ×A× (Ti - To)/L
Substituting the values into the equation
H = 0.04 × K × 1.4 ×(175-35)/4×10^-2
H= 196W.
The electric power = Heat current =196W
The electric power is the magnitude of heat current
The correct answer is: The voltage drop across 
is 45V.
Explanation:
First you need to find the total current of the circuit, which is:
According to Ohm's law:
V = I * R
I = V/R --- (1)
Where
R = Total Equivalent resistance of the circuit
V = Total Voltage = 90V
I -= Total Current
First let us find the equivalent resistance:
Now plug in the values in (1):
(1) => I = 90/30 = 3A
Now that we have current of the circuit, we can now find the voltage drop across = 15 ohms using:
Where
is the voltage drop across resistor 5;
Explanation:
First you need to find the total current of the circuit, which is:
According to Ohm's law:
V = I * R
I = V/R --- (1)
Where
R = Total Equivalent resistance of the circuit
V = Total Voltage = 90V
I -= Total Current
First let us find the equivalent resistance:
Now plug in the values in (1):
(1) => I = 90/30 = 3A
Now that we have current of the circuit, we can now find the voltage drop across
Where