Answer:
[tex]Q=-2.444*10^{-13}C[/tex]
Explanation:
Given data
Potential V= -2.2 V
Distance from charges r=1 mm=10⁻³m
Electric force constant k=9×10⁹N.m²/C²
To find
Sign and magnitude of a point charge
Solution
The electric potential of point charge is given as:
[tex]V=\frac{kQ}{r}\\ Q=\frac{Vr}{k}\\ Q=\frac{(-2.2)(10^{-3} )}{9*10^{9} }\\Q=-2.444*10^{-13}C[/tex]
