(A) The potential gradient is defined as the potential drop per unit length,given by $E = \frac{V}{L}$.From Ohm's law,$V = iR$,where $R = \rho \frac{L

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(A) The potential gradient is defined as the potential drop per unit length,given by $E = \frac{V}{L}$.From Ohm's law,$V = iR$,where $R = \rho \frac{L}{A}$.Substituting $R$ into the potential gradient formula:$E = \frac{i \rho L}{A L} = \frac{i \rho}{A}$.Given values: $i = 0.2 \, A$,$\rho = 40 \times 10^{-8} \, \Omega \, m$,and $A = 8 \times 10^{-6} \, m^2$.Substituting these values:$E = \frac{0.2 \times 40 \times 10^{-8}}{8 \times 10^{-6}}$.$E = \frac{8 \times 10^{-8}}{8 \times 10^{-6}} = 10^{-2} \, V/m$.

Class 12 Physics Electrostatic Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

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What is the potential gradient in a wire of resistivity $40 \times 10^{-8} \, \Omega \, m$ and cross-sectional area $8 \times 10^{-6} \, m^2$ when a current of $0.2 \, A$ flows through it?

A
$10^{-2} \, V/m$
B
$10^{-1} \, V/m$
C
$3.2 \times 10^{-2} \, V/m$
D
$1 \, V/m$

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Electrostatic Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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What is the potential gradient in a wire of resistivity $40 \times 10^{-8} \, \Omega \, m$ and cross-sectional area $8 \times 10^{-6} \, m^2$ when a current of $0.2 \, A$ flows through it?

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Two metal plates are separated by $2 \,cm$. The potentials of the plates are $-10 \,V$ and $+30 \,V$. The electric field between the two plates is: (in $\,V/m$)

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