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Two point charge +-q are placed on the a...

Two point charge `+-q` are placed on the axis at `x= -a` and `x= +-a`, as shown in fig.
(i) Plot the variation of E along the x-axis.
(ii) Plot the variation of E along the y-axis.

Text Solution

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We can divide the x-axis into three region:
Region I (left of the charges): The field due to the left charge dominates and acts towards-x direction.
Region II(between the charges): From `-altxlt0`, the field of left charge dominater and acts towads.+x direction. At x=0,
resulant electric field is zero. from `0ltxlta`, the field of right charge dominaters and acts towards `+x` direction


Variation of `vec(E)` in the y-axis: we can divide the y-axis in two regions:
Resion I (above y-axis): Electric field is along `+y` direction.

Resion II (below y-axis ): Electric field is along y direction. Net electric field at P at a distance y from origin is
`E_("net")=|E_(1)|sin theta+|E_(2)|sin theta=2|E_(1)|sin theta` as `|vecE_(1)|=|vecE_(2)|`
`=2(1)/(4 pi epsilon_(0)) (q)/((a^(2)+y^(2))) (y)/((a^(2)+y^(2))^(1//2))`
Hence, `E_("net")=(1)/(4 pi epsilon_(0)) (qy)/((a^(2)+y^(2))^(3//2))`
For maximum or minimum value of `E_("net")`
`(dE_("net"))/(dy)=0=(q)/(2 pi epsilon_(0))[y((3)/(2))(a^(2)+y^(2))^(-5//2)(2y)+(a^(2)+y^(2))^(-3//2)(1)]`
or, `y= +- (a)/(sqrt(2))`
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