Show that the maximum magnitude Emax of the electric field along the azies of a uniformly charged ring occurs at x=a/(2)^(1/2)
and has the value Q/(6(3)^(1/2)pieEoa^2)
using the fact that:
E=KeQx/(x^2 + a^2)^(3/2)
The solution manual shows that
for maximum , dE/dx=QKe[ (1/(x^2 + a^2)^(3/2) ) - (3x^2/(x^2 + a^2)^(5/2) ]
I don't understand how they arrive to that point, I figure before it was a derivative but I did one taking the E=KeQx/(x^2 + a^2)^(3/2) and I don't get that answer can some one show me step by step, thank you, I understand the rest of the problem. Could anyone help me, Thank you!
4 hours ago - 4 days left to answer.
and has the value Q/(6(3)^(1/2)pieEoa^2)
using the fact that:
E=KeQx/(x^2 + a^2)^(3/2)
The solution manual shows that
for maximum , dE/dx=QKe[ (1/(x^2 + a^2)^(3/2) ) - (3x^2/(x^2 + a^2)^(5/2) ]
I don't understand how they arrive to that point, I figure before it was a derivative but I did one taking the E=KeQx/(x^2 + a^2)^(3/2) and I don't get that answer can some one show me step by step, thank you, I understand the rest of the problem. Could anyone help me, Thank you!
4 hours ago - 4 days left to answer.