Given Information:
Electric field = E = 180,000 N/C
Distance = r = 1.50 cm = 0.015 m
Required Information:
Charge = q = ?
Answer:
charge = q = 4.5x10⁻⁹ C
Explanation:
The electric field is given by
E = kq/r²
Where k is the coulomb constant k = 9x10⁹ Nm²/C²
q = Er²/k
q = 180,000*(0.015)²/9x10⁹
q = 4.5x10⁻⁹ C
Therefore, the charge is 4.5 nC
(98.6 - 98.2)ºF = .4ºF
.4ºF* 5ºC/9ºF = 0.222 ºC
60
so you take 120÷2 wires
hi your pinterest I'd or Twitter I'd pls.