Answer:
[tex]Q = 6.69 \times 10^{-8} C[/tex]
Explanation:
Electric field due to a finite wire at some distance "x" from it on its perpendicular bisector is given by the formula
[tex]E = \frac{k\lambda}{r}(sin\theta_1 + sin\theta_2)[/tex]
so here we will have
[tex]\theta_1 = \theta_2 = tan^{-1}(\frac{5}{4})[/tex]
[tex]\theta_1 = 51.3 ^o[/tex]
now we know that
F = qE
we will have
[tex]940 \times 10^{-6} = (4 \times 10^{-9})(\frac{(9\times 10^9)\lambda}{0.04}(2sin51.3))[/tex]
so we will have
[tex]\lambda = 6.69 \times 10^{-7} C/m[/tex]
so total charge on the rod is given as
[tex]Q = \lambda L[/tex]
[tex]Q = (6.69 \times 10^{-7})(0.10)[/tex]
[tex]Q = 6.69 \times 10^{-8} C[/tex]
The total charge on the rod at the given electric field strength is [tex]6.68 \times 10^{-8} \ C[/tex].
The given parameters;
The electric field due to the total charge on the rod is calculated as follows;
[tex]E = \frac{KQ}{ry} \\\\y = \sqrt{r^2 + \frac{L^2}{4} } \\\\y = \sqrt{(0.04)^2 + \frac{(0.1^2)}{4} } \\\\y = 0.064 \ m\\\\E = \frac{kQ}{0.064r}[/tex]
The force experienced by the charged bead is calculated as;
[tex]F = qE\\\\F = \frac{kqQ}{0.0.064 r} \\\\Q = \frac{0.064 rF}{kq} \\\\Q = \frac{0.064 \times 0.04 \times 940 \times 10^{-6}}{9\times 10^9 \times 4\times 10^{-9}} \\\\ Q= 6.68 \times 10^{-8} \ C[/tex]
Thus, the total charge on the rod at the given electric field strength is [tex]6.68 \times 10^{-8} \ C[/tex].
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