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To find the potential (V) at (x = 0.5 , text{mm}), given that the potential (V) at (x = 1 , text{mm}) is (2 , text{V}), and the volume charge density (rho_v) is (-10^6 varepsilon_0), constant throughout the free space region between (x = 0) and (x = 1 , text{mm}), we’ll use the concept of electric field and potential in a region with constant volume charge density.
Given data:
– (rho_v = -10^6 varepsilon_0) (Volume charge density, where (varepsilon_0) is the permittivity of free space (left(varepsilon_0 = 8.85 times 10^{-12} , text{F/m}right))).
– (V(x = 1 , text{mm}) = 2 , text{V})
– (V(x = 0) = 0 , text{V}) (Since it’s mentioned that the potential is zero at (x = 0).)
– (x = 0.5 , text{mm}) is the point where we need to find the potential.
To find the potential at (x = 0.
d
Explanation: Del2
(V) = -ρv/εo= +106
On integrating twice with respect to x, V = 106. (x2
/2) + C1x + C2.
Substitute the boundary conditions, x = 0, V = 0 and x = 1mm, V = 2V in V,
C1 = 1500 and C2 = 0. At x = 0.5mm, we get, V = 0.875V