Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.1×109 electrons from one disk to the
other causes the electric field strength between them to be 1.6×105 N/C .
2 answers:
Answer:
Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.1×109 electrons from one disk to the other causes the electric field strength between them to be 1.6×105 N/C. What are the diameters of the disks?
Explanation:
Check attachment for solution
Complete question:
Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.1 × 10⁹ electrons from one disk to the other causes the electric field strength between them to be 1.6 × 10⁵ N/C . What are the diameters of the disks ?
Answer:
The diameter of the disks is 0.0174 m
Explanation:
Given;
charge, q on each sphere = 2.1 × 10⁹ x 1.6 x 10⁻¹⁹ C = 3.36 x 10⁻¹⁰ C
Electric field strength due to charged spheres, E = 1.6 × 10⁵ N/C
E = V/d
Capacitance is given as;
C = εA/d
The charge on a capacitor is given as;
Q = CV
Q =
But, A = πd²/4
Q = E(πd²/4)ε
where;
d is the diameter of the disks
ε is permittivity of free space = 8.854 x 10⁻¹² F/m
Substitute the given values and solve for "d"
Therefore, the diameter of the disks is 0.0174 m
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Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A =
where d is the diameter of the wire
1.613 x 10^-7 =
6.448 x 10^-7 = 3.142 x
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>