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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