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".
Answer:
The toy's energy is 18 J.
Explanation:
We have, a 4 kg toy is lifted off the ground and falls at 3 m/s. It is required to find toy's energy.
The toy will have kinetic energy due to its motion. The energy is given by :

So, the toy's energy is 18 J.
For the electric field is measured for points at distances r,Electric field is mathematically given as
E=19.9*10^{-5}c/m^3
<h3>What is the
Electric field?</h3>
Generally, the equation for the electric field uniform charged is mathematically given as

Therefore
E=lr/3E0=3Ee0/r
Therefore
E=3*6*10^4*8.85*10^{-12}
E=19.9*10^{-5}c/m^3
In conclusion, E=19.9*10^{-5}c/m^3
E=19.9*10^{-5}c/m^3
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CQ
Complete question attached below
The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
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