Common transparent tape becomes charged when pulled from a dispenser. If one pie

Question

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 in C if electrostatic force is great enough to support the weight of a 0.65 mg piece of tape held 0.95 cm above another. Assume both pieces of tape have the same charge. What force must Coulomb's law overcome to prevent the tape from falling?

Explanation / Answer

Given a = 2.5 cm = 2.5 x10 -2 m

         b = 1.2 cm = 1.2 x10 -2 m

Force on B due to A is F = K(qA)(qB) / a 2

Where K = Coulomb's constant

               = 8.99 x10 9 Nm 2/C 2

          qA = -1 nC = -1 x10 -9 C

          qB = -2 nC = -2 x10 -9 C

Substitute values you get , F =(8.99 x10 9 )(1 x10 -9 ) (2x10 -9 ) / (2.5 x10 -2 )

                                          = 7.192 x10 -7 N

Where F is along AB direction.i.e., downward

Force on B due to C is F ' = K(qC)(qB) / b 2

Where K = Coulomb's constant

               = 8.99 x10 9 Nm 2/C 2

          qC = +2 nC = 2 x10 -9 C

          qB = -2 nC = -2 x10 -9 C

Substitute values you get , F ' =(8.99 x10 9 )(2 x10 -9 ) (2x10 -9 ) / (1.2 x10 -2 )

                                          = 2.9966 x10 -6 N

Where F ' is along BC direction.i.e., downward

Therefore net electric force on B is = F + F '

                                                  = 3.715x10 -6 N

(b). The force is directed downward

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