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Show that if x>>a...
I've got the following force generated by an electric field
[tex]{F_x} = - KQq\frac{1}{{x\sqrt {{x^2} + {a^2}} }}[/tex]
[tex]{F_y} = \frac{{KQq}}{a}\left( {\frac{1}{x} - \frac{1}{{\sqrt {{x^2} + {a^2}} }}} \right)[/tex]
I need to show that when x>>a:
[tex]{F_x} = - \frac{{KQq}}{{{x^2}}}[/tex]
[tex]{F_y} = \frac{{KQqa}}{{2{x^3}}}[/tex]
I think I'm on the right track but I'm stuck here:
[tex]\frac{1}{{\sqrt {{x^2} + {a^2}} }} = \frac{1}{x}{\left( {1 + \frac{{{a^2}}}{{{x^2}}}} \right)^{ - \frac{1}{2}}}[/tex]
Homework Statement
I've got the following force generated by an electric field
[tex]{F_x} = - KQq\frac{1}{{x\sqrt {{x^2} + {a^2}} }}[/tex]
[tex]{F_y} = \frac{{KQq}}{a}\left( {\frac{1}{x} - \frac{1}{{\sqrt {{x^2} + {a^2}} }}} \right)[/tex]
Homework Equations
I need to show that when x>>a:
[tex]{F_x} = - \frac{{KQq}}{{{x^2}}}[/tex]
[tex]{F_y} = \frac{{KQqa}}{{2{x^3}}}[/tex]
The Attempt at a Solution
I think I'm on the right track but I'm stuck here:
[tex]\frac{1}{{\sqrt {{x^2} + {a^2}} }} = \frac{1}{x}{\left( {1 + \frac{{{a^2}}}{{{x^2}}}} \right)^{ - \frac{1}{2}}}[/tex]