Question

A stick of length 6.2 cm has charged globes on either end, one charged to 16 μC, while the other one is charged to 1μC. The rod is placed perpendicular to a 1174 N/C field. What is the force on the stick?

What is the polarization of the bar?
How much torque does it feel?

Answer 1

Net force = 0.0191 N

T=0.00054 N.m

p=99.2 μC.cm

Step-by-step explanation:

Given that

L= 6.2 cm

q₁= 16μC

q₂=1μC

E= 1174 N/C

We know that

F = q E

F₁ = q₁ E

`F_1=16* 10^(-6)* 1174\ N`

F₁ = 0.018 N

F₂=q₂ E

`F_2=1* 10^(-6)* 1174\ N`

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