A thin plastic rod of length 2.8 m is rubbed all over with wool, and acquires a charge of 88 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 15 cm from the midpoint of the rod. Do the calculation two ways, first using the exact formula for a rod of any length, and second using the approximate formula for a long rod.

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]

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