Answer:
C. 5.6 × 10^11 N/C
Explanation:
The electric field [tex]E[/tex] at a distance [tex]R[/tex] from a charge [tex]Q[/tex] is given by
[tex]E = k\dfrac{Q}{R^2}[/tex]
where [tex]k = 9*10^9Nm/C[/tex] is the coulomb's constant.
Now, in our case
[tex]R = 0.0075m[/tex]
[tex]Q = 0.0035C[/tex];
therefore,
[tex]E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}[/tex]
[tex]\boxed{E = 5.6*10^{11}N/C.}[/tex]
which is choice C from the options given (at least it resembles it).
