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2 months ago

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

Physics

1 answer:

serg [3.5K]2 months ago

6 0

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k represents Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ denotes the charge magnitude on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

$e=-1.6\cdot 10^{-19}C$

$q_2 = Ne$

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The overall charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

$e=-1.6\cdot 10^{-19}C$

$q_1 = -Ne$

$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

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