The field potential in a certain region of space depends only on the x coordinate as φ φ = –ax 3 + b, where a and b are constants. Find the distribution of the space charge ρ(x).
Text Solution
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(6aε 0 x)
Given that field potential is variable only in x direction. so for Given value of x (also we can conclude that electric field is parallel to x-axis) Potential is constant in Y and Z direction. Now, by taking a small volume in space at distance x. Cross section area of this element is A and width is dx [Given φ φ = ax 3 + b]
Electric field 
= – 
= –3ax 2 
Now electric field is constant for a particular value of x and it is parallel to area vector of this small elemental volume.
By Guess theorem:
EA = 
⇒ A dE = 
A.6axdx = 
⇒ 
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