Respuesta :
Answer:
Exact formula E=3749.9 N/C
Approximate formula E=3771.43 N/C
Explanation:
Given that
L= 2.8 m
q= 88 nC
a= 15 cm
By using exact formula:
We know that due to charged rod given as
[tex]E=\dfrac{Kq}{a}\times \dfrac{1}{\sqrt{a^2+\dfrac{L^2}{4}}}[/tex]
Now by putting the values
[tex]E=\dfrac{9\times 10^9\times 88\times 10^{-9}}{0.15}\dfrac{1}{\sqrt{0.15^2+\dfrac{2.8^2}{4}}}\ N/C[/tex]
E=3749.9 N/C
By approximate formula:
[tex]a^2+\dfrac{L^2}{4}=\dfrac{L^2}{4}[/tex]
because L is so long as compare to a.
[tex]E=\dfrac{Kq}{a}\times \dfrac{1}{\sqrt{\dfrac{L^2}{4}}}[/tex]
[tex]E=\dfrac{9\times 10^9\times 88\times 10^{-9}}{0.15}\dfrac{1}{\sqrt{\dfrac{2.8^2}{4}}}\ N/C[/tex]
E=3771.43 N/C
Answer:
(I). The electric field when we using the exact formula is [tex]37.5\times10^{2}\ N/C[/tex]
(II). The electric field when we using the approximate formula is [tex]37.7\times10^{2}\ N/C[/tex]
Explanation:
Given that,
length l= 2.8 m
Charge = 88 nC
Distance from midpoint of the rod = 15 cm
(a). We need to calculate the electric field
Using the exact formula for a rod of any length
[tex]E=\dfrac{kQ}{a}(\dfrac{1}{\sqrt{a^2+(\dfrac{L}{2})^2}})[/tex]
Where, Q = charge
l = length
a = distance
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times88\times10^{-9}}{15\times10^{-2}}(\dfrac{1}{\sqrt{(15\times10^{-2})^2+(\dfrac{2.8}{2})^2}})[/tex]
[tex]E=37.5\times10^{2}\ N/C[/tex]
(b). We need to calculate the electric field
Using the approximate formula for a long rod
[tex]E=\dfrac{kQ}{a}(\dfrac{1}{\sqrt{a^2+(\dfrac{L}{2})^2}})[/tex]
a<<L, then [tex]a^2+(\dfrac{L}{2})^2=(\dfrac{L}{2})^2[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times88\times10^{-9}}{15\times10^{-2}}(\dfrac{1}{\sqrt{(\dfrac{2.8}{2})^2}})[/tex]
[tex]E=37.7\times10^{2}\ N/C[/tex]
Hence, (I). The electric field when we using the exact formula is [tex]37.5\times10^{2}\ N/C[/tex]
(II). The electric field when we using the approximate formula is [tex]37.7\times10^{2}\ N/C[/tex]
