A line charge exists on A and B is 4m- - 5m then the potential difference VAB - VA - VB Z-axis with -1 - 10-92 C-m If A is 2m- -2- 0

V_AB = ∫_B^AE.dt Where, nbsp; E = ρ l2 πϵ _0 râ_r V_AB = ∫_B^A10^ 92(2 πϵ _0 r)dr = 10^ 9× 36 π nbsp;4 π× 10^ 9∫_B^A1r dr = 9 [ln r]^2_4 V_AB = 6.24 ...

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Electric Potential

A line charge...

Question

A line charge exists on A and B is $(4 m, π, 5 m)$ then the potential difference $V_{A B} = V_{A} - V_{B}$
Z-axis with $ρ_{1} = \frac{10^{- 9}}{2} (C / m)$ If $A$ is $(2 m, \frac{π}{2}, 0)$

5.64 V

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6.24 V

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8.91 V

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5.23 V

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Solution

The correct option is B 6.24 V
$V_{A B} = - \int_{B}^{A} \to E . \to d t$

Where, $\to E = \frac{ρ l}{2 π ϵ_{0} r} {^a}_{r}$

$V_{A B} = - \int_{B}^{A} \frac{10^{- 9}}{2 (2 π ϵ_{0} r)} d r = \frac{- 10^{- 9} \times 36 π}{4 π \times 10^{- 9}} \int_{B}^{A} \frac{1}{r} d r$

$= - 9 [ln r]_{4}^{2}$

$V_{A B} = 6.24 V$

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A line charge exists on A and B is (4m,π,5m) then the potential difference VAB=VA−VB Z-axis with ρ1=10−92(C/m) If A is (2m,π2,0)

The correct option is B 6.24 VVAB=−∫AB→E.→dt Where, →E=ρl2πϵ0r^ar VAB=−∫AB10−92(2πϵ0r)dr=−10−9×36π4π×10−9∫AB1rdr =−9[lnr]24 VAB=6.24 V

A line charge exists on A and B is $(4 m, π, 5 m)$ then the potential difference $V_{A B} = V_{A} - V_{B}$
Z-axis with $ρ_{1} = \frac{10^{- 9}}{2} (C / m)$ If $A$ is $(2 m, \frac{π}{2}, 0)$