Answer:
Charge, q = 3.48 μC
Explanation:
It is given that,
Electric field, [tex]E=1.25\times 10^5\ N/C[/tex]
Distance form positive charge, r = 0.5 meters
We need to find the charge. It is given by the formula as follows :
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]q=\dfrac{Er^2}{k}[/tex]
k = electrostatic constant
[tex]q=\dfrac{1.25\times 10^5\ N/C\times (0.5\ m)^2}{9\times 10^9}[/tex]
q = 0.00000347 C
or
[tex]q=3.48\times 10^{-6}\ C[/tex]
[tex]q=3.48\ \mu C[/tex]
So, the charge is 3.48 μC. Hence, this is the required solution.
