Coulomb’s Law describes the force between two (and only two!) charged objects. S
ID: 1404890 • Letter: C
Question
Coulomb’s Law describes the force between two (and only two!) charged objects. Suppose Object 1 has a charge of +5.46 C and Object 2 has a charge of +14.2 C Object 2 is located a distance of 0.04460 m to the right of Object 1 (Yes — please break down and draw a quick sketch!).
a) What is the magnitude and direction of the electrical force on Object 2 due to Object 1?
b)What is the magnitude and direction of the electrical force on Object 1 due to Object 2?
c)You now move Object 2 such that it is located a distance of 0.08920 m to the right of Object 1 (twice the distance as before). What is the magnitude and direction of the electrical force on Object 2 due to Object 1?
d)The new force on Object 1 due to Object 2 (at a separation of 0.08920 m) is ____ times the original force on Object 1 due to Object 2 (at a separation of 0.04460 m)? (Provide a numerical answer to fill in the blank.)
Explanation / Answer
a)
force between two charged particles is given as
F = k *q1*q2/d^2
Now, for force on object 2 due to 1
force on object 2 due to 1 = 9*10^9 * 5.46 *10^-6 * 14.2 *10^-6/.0446^2
force on object 2 due to 1 = 350.8 N
as the chagres are of same sign , they repel hence ,
force on object 2 due to 1 is 350.8 N to the right
b)
for force on object 1 due to 2
force on object 1 due to 2 = 9*10^9 * 5.46 *10^-6 * 14.2 *10^-6/.0446^2
force on object 1 due to 2 = 350.8 N
as the chagres are of same sign , they repel hence ,
force on object 1 due to 2is 350.8 N to the left
c)
for twice the distance ,
Now, for force on object 2 due to 1
force on object 2 due to 1 = 9*10^9 * 5.46 *10^-6 * 14.2 *10^-6/.0892^2
force on object 2 due to 1 = 87.7 N
as the chagres are of same sign , they repel hence ,
force on object 2 due to 1 is 87.7 N to the right
d)
as the distance is doubled , the force become (1/2)^2 = 1/4 times
hence , the new force is 0.25 times the previous force