Question

One point charge has a magnitude 5.4x10-7 C. A second charge 0.25 meters away has a magnitude of 1.1x10-17 C. What is the electric force magnitude of one charge on the other?

Answer 1

The electric force on the charges will be equal and opposite in nature and the magnitude will be 8.5536 × 10⁻¹³ Newton.

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

In question we have given

value of q1 = 5.4 x 10⁻⁷ C

value of q2 = 1.1 x 10⁻¹⁷ C

distance between the (r) = 0.25 m

Applying the formula

F = ( K x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 10⁹ ) x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 5.4 × 1.1) × ( 10⁹ × 10⁻⁷ x 10⁻¹⁷) )/ ( 6.25 × 10⁻² )

F = ( 53.46 × 10⁻¹⁵) / ( 6.25 × 10⁻² )

F = 8.5536 × 10⁻¹³ Newton

Electrostatic force = 8.5536 × 10⁻¹³ Newton

So, The point charges are possessing equal and opposite electrostatic force of magnitude 8.5536 × 10⁻¹³ Newton.

Learn more about Electrostatic Force here:

brainly.com/question/23121845

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