Question

Two object accumulated a charge of 4.5 μC and another a charge of 2. 8 μC. These two objects are separated by a distance of 2.5 cm. Calculate the magnitude of the electric force between these two objects.

Answer 1

Answer: 181.23 C

Step-by-step explanation:

According to Coulomb's Law:

`F_(E)= K(q_(1).q_(2))/(d^(2))`

Where:

`F_(E)` is the electrostatic force

`K=8.99(10)^(9) Nm^(2)/C^(2)` is the Coulomb's constant

`q_(1)=4.5 \mu C=4.5(10)^(-6) C` is the charge of object 1

`q_(2)=2.8 \mu C=2.8(10)^(-6) C` is the charge of object 2

`d=2.5 cm (1 m)/(100 cm)=0.025 m` is the separation distance between the charges

`F_(E)=8.99(10)^(9) Nm^(2)/C^(2)((4.5(10)^(-6) C)(2.8(10)^(-6) C))/((0.025 m)^(2))`

`F_(E)=181.23 C`