Answer:
[tex]1\times 10^{8} N/C[/tex]
Explanation:
According to the gauss law
As we know that
Electric field is
[tex]E = -k\frac{q}{r^2}[/tex]
where,
k = column constant = [tex]9 \times 10 ^{9}\ N. \frac{m^2}{c^2}[/tex]
q = charge
r = distance from the sphere center
For computing the magnitude of e first we have to need to find out the charge outside of sphere which is
[tex]q = -\frac{Er^2}{k}[/tex]
[tex]q = -\frac{1 \times 10^{6} \frac{N}{C} (25m)^2}{9 \times 10 ^{9}\ N. \frac{m^2}{c^2}}[/tex]
q = -0.07 C
Now we have to find the electric field
[tex]E = k\frac{q}{r^2}[/tex]
The r is 2.5m but in question it is given 5m
So,
Electric field is
[tex]E = 9 \times 10^{9} N . \frac{m^2 \times 0.07 C}{C^2 (2.5 m)^2}[/tex]
[tex]E = 9 \times 10^{9} N. \frac{m^2 \times 0.07 C} {C^2 (2.5m)^2}[/tex]
[tex]= 1\times 10^{8} N/C[/tex]