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A 7.80 nC charge is located 1.69 m from a 4.54 nC point charge.

(a) Find the magnitude of the electrostatic force that one charge exerts on the other.

Answer :

The electric force acting on a 7.80 nc charge that is located 1.69 m from a 4.54 nc point charge is: 1.115*10^-7 N

To solve this exercise the electric force formula and the procedure we will use is:

F = (k * q1 * q2)/r²

Where:

  • F = electric force
  • k = coulomb constant
  • q1 = charge 1
  • q2 = charge 2
  • r = separation distance of the charges

Information about the problem:

  • q1= 7.80 nc
  • q2= 4.54 nc
  • r = 1.69 m
  • F =?
  • k= 9 *10^9 N*m²/C²
  • 1 C = 1*10^ 9 nC

By converting the the values of the charges (q1) and (q2) from (nC) to (C) and we have that:

q1= 7.80 nC * 1 C/ 1*10^ 9 nC

q1= 7.8*10^ -9 C

q2= 4.54 nC * 1 C/ 1*10^ 9 nC

q2= 4.54 *10^ -9 C

We apply the electric force formula we have:

F = (k * q1 * q2)/r²

F = [(9 *10^9 N*m²/C² * (7.8*10^ -9 C) * (4.54 *10^ -9 C)]/ (1.69 m)²

F = 3.187*10^-7 N*m² /2.8561 m²

F = 1.115*10^-7 N

What is electric force?

In physics the electric force is the force that attracts or repels two charges (q) separated at a distance called (r), this is expressed in the international system of units in Newton.

Learn more about electric force at: brainly.com/question/25923373

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