Question

3. (a) Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a9.00 mg piece of tape held 1.10 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)

5. (a) Two point charges totaling 8.00 µC exert a repulsive force of 0.100 N on one another when separated by 0.553 m. What is the charge on each?
smallest charge

largest charge


5(b). What is the charge on each if the force is attractive?
smallest charge

largest charge

Answer 1

3(a)

by balencing the force,

Fg=mg

and

Fe=k*q1*q2/r^2

now,

Fg=Fe

mg=k*q1*q2/r^2

let, q1=q2=q

mg=k*q^2/r^2

(9*10^-3*9.8)=(9*10^9*q^2)/(1.1*10^-2)^2

===>

q=3.44*10^-8 C -----is answer

5(a)

given,

(q1+q2)=8uC=8*10^-6 c ------(1)

F=0.1 N

r=0.553 M

now,

F=k*q1*q2/r^2

0.1=9*10^9*q1*q2/0.553^2

===>

q1*q2=3.4*10^-12 C^2 -------(2)

from eqn no (1) and (2)

(q1-q2)=sqrt[(q1+q2)^2-(4*q1*q2)]

(q1-q2)=sqrt[(8*10^-6)^2-(4*3.4*10^-12)]

(q1-q2)=7.1*10^-6 C -----(3)

and

(q1+q2)=8*10^-6 c ------(1)

by solving above two equation (1) and (3)

q1=7.55*10^-6 C

q1=7.55 uC

and

q2=0.45*10^-6 C

q2=0.45 uC

hence

smallest charge q2=0.45 uC

largest charge q1=7.55 uC

5(b)

given,

(q1-q2)=8uC=8*10^-6 c ------(1)

F=k*q1*q2/r^2

0.1=9*10^9*q1*q2/0.553^2

===>

q1*q2=3.4*10^-12 C^2 -------(2)

from eqn no (1) and (2)

(q1+q2)=sqrt[(q1-q2)^2+(4*q1*q2)]

(q1+q2)=sqrt[(8*10^-6)^2+(4*3.4*10^-12)]

(q1+q2)=8.81*10^-6 C -----(3)

and

(q1-q2)=8.8*10^-6 c ------(1)

by solving above two equation (1) and (3)

q1=8.805*10^-6 C

q1=8.805 uC

and

q2=0.005*10^-6 C

q2=0.005 uC

hence

smallest charge q2=0.005 uC

largest charge q1=8.805 uC