The correct field line would be A.
Answer:
E) 6.5 A
Explanation:
Given that
L = 40 m H
C= 1.2 m F
Maximum charge on capacitor ,Q= 45 m C
The maximum current I given as
I = Q.ω
ω =angular frequency

By putting the values


ω = 144.33 rad⁻¹
Maximum current
I = 45 x 10⁻³ x 144.33 A
I= 6.49 A
I = 6.5 A
E) 6.5 A
An electron has a negative charge
Answer:
110.87 dB
Explanation:
(I got it right on Acellus)
I= P/4(pi)r^2 = 60/4(pi)6.25^2
60/4(pi)6.25^2=0.12223
B=10log(I/Io)
B=10log(0.12223/1*10^-12) = 110.87 dB
111 in sigfigs