Answer:
The magnitude of rate of change of electric field is 
.
Explanation:
Given that,
Radius of the cylindrical region contains a uniform electric field along the cylinder axis, r = 1.2 m
Total displacement current through a cross section of the region, 
We need to find the rate of change of electric field. Its is given by the formula as follows :
So, the magnitude of rate of change of electric field is .
Answer:
r = 0.114 m
Explanation:
To find the speed of the proton, from conservation of energy, we know that
KE = PE
Thus, we have;
(1/2)mv² = qV
Where;
V is potential difference = 1kv = 1000V
q is charge on proton which has a value of 1.6 x 10^(-19) C
m is mass of proton with a constant value of 1.67 x 10^(-27) kg
Let's make the velocity v the subject;
v =√(2qV/m)
v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))
v = 4.377 x 10^(5) m/s
Now to calculate the radius of the circular motion of charge we know that;
F = mv²/r = qvB
Thus, mv²/r = qvB
Divide both sides by v;
mv/r = qB
Thus, r = mv/qB
Value of B from question is 0.04T
Thus,
r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)
r = 0.114 m
r = 8.76 m
Answer:
False
Explanation:
The second you let go its gonna release kinetic energy that's why it's potential
3.1 the only reason i know this is cause i got it wrong
<span>A decrease in the overall volume of gases namely hydrogen would prevent nuclear fusion in a nebula.</span>
