Electric potential in a region is given by ph | vedclass

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(A) The electric field $\vec{E}$ is related to the electric potential $\phi$ by the relation $\vec{E} = -
abla \phi$.Since the potential $\phi$ depen

Vedclass · Accepted Answer

$0.40 \ Vm^{-1} \hat{i}$

Vedclass · Answer

(A) The electric field $\vec{E}$ is related to the electric potential $\phi$ by the relation $\vec{E} = -
abla \phi$.Since the potential $\phi$ depends only on $x$,the electric field is given by $\vec{E} = -\frac{\partial \phi}{\partial x} \hat{i}$.Given $\phi(x) = \phi_0 \frac{x_0}{x} = (8 \ V)(5 \ m) \frac{1}{x} = \frac{40}{x} \ V \cdot m$.Differentiating with respect to $x$: $\frac{\partial \phi}{\partial x} = 40 \frac{d}{dx}(x^{-1}) = 40 (-x^{-2}) = -\frac{40}{x^2}$.Therefore,$\vec{E} = -(-\frac{40}{x^2}) \hat{i} = \frac{40}{x^2} \hat{i} \ V/m$.At the point $(10 \ m, 5 \ m, 5 \ m)$,the $x$-coordinate is $10 \ m$.Substituting $x = 10 \ m$ into the expression for $\vec{E}$: $\vec{E} = \frac{40}{10^2} \hat{i} = \frac{40}{100} \hat{i} = 0.40 \ Vm^{-1} \hat{i}$.

Electric potential in a region is given by $\phi(x, y, z) = \phi_0 \frac{x_0}{x}$; where $x_0 = 5 \ m$ and $\phi_0 = 8 \ V$. Find the electric field at $(10 \ m, 5 \ m, 5 \ m)$.

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