Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
Answer:
5.63
Explanation:
Divide mass by volume, which will give you 5.63. For instance, 45 divide by 8 what is it? find out
Answer:
1.99 parsecs.
Explanation:
We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.
We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.
Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.
Answer: I = 111.69 pA
Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.
The hall voltage of a semiconductor sensor is given below as
V = I×B/qnd
Where V = hall voltage = 1.5mV =1.5/1000=0.0015V
I = current =?,
n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3
q = magnitude of an electronic charge=1.609×10^-19c
B = strength of magnetic field = 5T
d = thickness of sensor = 0.8mm = 0.0008m
By slotting in the parameters, we have that
0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008
0.0015 = I×5/7.446×10^-8
I = (0.0015 × 7.446×10^-8)/5
I = 111.69*10^(-12)
I = 111.69 pA
