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प्रश्न
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
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उत्तर
Take a long thin wire XY (as shown in the figure) of uniform linear charge density `lambda`.
Consider a point A at a perpendicular distance l from the mid-point O of the wire, as shown in the following figure.
Let E be the electric field at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
∴ `"q" = lambda "dx"`
Electric field due to the piece,
`"dE" = 1/(4piin_0)(lambda "dx")/(("AZ")^2)`
However, `"AZ" = sqrt(("l"^2 + "x"^2))`
`"dE"= ( lambda "dx")/(4piin_0("l"^2 + "x"^2))`
The electric field is resolved into two rectangular components. `"dE"cos theta` is the perpendicular component and `"dE"sintheta` is the parallel component.
When the whole wire is considered, the component `"dE"sintheta` is cancelled.
Only the perpendicular component `"dE"cos theta` affects point A.
Hence, effective electric field at point A due to the element dx is dE1.
∴ `"dE"_1 = (lambda"dx" costheta)/(4piin_0("x"^2 + "l"^2))` .............(1)
In `triangle"AZO",`
`tantheta = "x"/"l"`
`"x" = "l" tantheta` ............(2)
On differentiating equation (2), we obtain `("dx")/("d" theta)= "l"sec^2theta`
`"dx" = "l"sec^2theta "d"theta` ........(3)
From equation (2),
`"x"^2 + "l"^2 = "l"^2 + "l"^2 tan theta`
∴ `"l"^2(1 + tan^2 theta) = "l"^2sec^2theta`
∴ `"x"^2 + "l"^2 = "l"^2sin^2theta` ........(4)
Putting equations (3) and (4) in equation (1), we obtain
∴ `"dE"_1= ( lambda "l" sec^2 "d" theta)/(4piin_0"l"^2sec^2 theta) xx costheta`
∴ `"dE"_1 = (lambda cos theta "d" theta)/(4 pi in_0 "l")` ..............(5)
The wire is so long that `theta` tends from `-pi/2` to `+pi/2`.
By integrating equation (5), we obtain the value of field E1 as,
`int_(-pi/2)^(pi/2) "dE"_1 = int_(-pi/2)^(pi/2)lambda/(4piin_0"l") cos theta "d" theta`
`"E"_1 = lambda/(4piin_0"l")[sin theta]_(-pi/2)^(pi/2)`
= `lambda/(4 pi in_0"l") xx 2`
`"E"_1 = lambda/(2piin_0"l")`
Therefore, the electric field due to long wire is `lambda/(2piin_0"l")`.
