Answer:
6908316.619 N/C
25087609.3949 N/C
6652357.02259 N/C
690831.6619 N/C
Explanation:
x = Distance from the ring
R = Radius of ring = 10 cm
q = Charge = 78 μC
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
Electric field at a point x is from a ring given by
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}[/tex]
For 1 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.01}{(0.01^2+0.1^2)^{1.5}}\\\Rightarrow E=6908316.619\ N/C[/tex]
The electric field is 6908316.619 N/C
For 5 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.05}{(0.05^2+0.1^2)^{1.5}}\\\Rightarrow E=25087609.3949\ N/C[/tex]
The electric field is 25087609.3949 N/C
For 30 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 0.3}{(0.3^2+0.1^2)^{1.5}}\\\Rightarrow E=6652357.02259\ N/C[/tex]
The electric field is 6652357.02259 N/C
For 100 cm
[tex]E=\dfrac{kqx}{(x^2+r^2)^{1.5}}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 78\times 10^{-6}\times 1}{(1^2+0.1^2)^{1.5}}\\\Rightarrow E=690831.6619\ N/C[/tex]
The electric field is 690831.6619 N/C
