350. If a potential V is 2V at x = 1mm and is zero at x=0 and volume charge density is -106εo, constant throughout the free space region between x = 0 and x = 1mm. Calculate V at x = 0.5mm.
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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.
d
See lessExplanation: 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.
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.
D
See lessExplanation: 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.
Answer: 0.875V 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.
Answer: 0.875V
See lessExplanation: 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.
Answer: 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.
Answer: d
See lessExplanation: 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.
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.
d)
See lessExplanation: 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.