Answer:
(A) E = 0.84 N/C
(B) r = 4.74 meters
Explanation:
Given that,
Charge, [tex]q=-5\ nC=-5\times 10^{-9}\ C[/tex]
(a) The magnitude of electric field at a point 7.3 meters from this charge is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]E=\dfrac{9\times 10^9\times 5\times 10^{-9}}{(7.3)^2}[/tex]
E = 0.84 N/C
We know that electric field due to negative charge is inwards. So, the direction of electric field is towards the charge.
(b) Let at a distance of d meters the electric field have a magnitude of 2 N/C. It is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]r=\sqrt{\dfrac{kq}{E}}[/tex]
[tex]r=\sqrt{\dfrac{9\times 10^9\times 5\times 10^{-9}}{2}}[/tex]
r = 4.74 meters
Hence, this is the required solution.
