Answer:
Net force = 0.0191 N
T=0.00054 N.m
p=99.2 μC.cm
Explanation:
Given that
L= 6.2 cm
q₁= 16μC
q₂=1μC
E= 1174 N/C
We know that
F = q E
F₁ = q₁ E
[tex]F_1=16\times 10^{-6}\times 1174\ N[/tex]
F₁ = 0.018 N
F₂=q₂ E
[tex]F_2=1\times 10^{-6}\times 1174\ N[/tex]
F₂=0.0011 N
Net force = F₁+F₂
Net force = 0.0191 N
Torque ,T
T = (F₁-F₂) L/2
T = (0.018 - 0.0011 ) x 0.032
T=0.00054 N.m
Polarization ,p
p= q d
p= 16 x 6.2 μC.cm
p=99.2 μC.cm
