Question

Electric potential due to a space is given by \[ \phi(x,y,z)=\phi_0\frac{x_0}{x} \] when \(x_0=5\,m\) and \(\phi_0=8\,V\). Find the electric field at \((10\,m,5\,m,5\,m)\)

• A. \(0.40\,Vm^{-1}\,\hat{i}\)
• B. \(-0.40\,Vm^{-1}\,\hat{i}\)
• C. \(4.0\,Vm^{-1}\,\hat{i}\)
• D. \(-4.0\,Vm^{-1}\,\hat{i}\)

Hint:

To find electric field from electric potential, use \[ \vec{E}=-\nabla \phi \] For one-dimensional potentials depending only on \(x\), \[ E_x=-\frac{dV}{dx} \]

Answer 1

The Correct Option is A

Step 1: Substitute the given constants.
The potential is $\phi(x,y,z) = \phi_0 \frac{x_0}{x}$. With $\phi_0 = 8\,\text{V}$ and $x_0 = 5\,\text{m}$:
\[ \phi = \frac{8 \times 5}{x} = \frac{40}{x} \]

Step 2: Recognise that the potential depends only on x.
Since $\phi$ does not depend on $y$ or $z$, the electric field has no $y$ or $z$ components.
Only $E_x$ is non-zero.

Step 3: Use the relation $E_x = -d\phi/dx$.
\[ E_x = -\frac{d\phi}{dx} = -\frac{d}{dx}\left(\frac{40}{x}\right) = -\left(-\frac{40}{x^2}\right) = \frac{40}{x^2} \]

Step 4: Evaluate at the given point $(10\,\text{m},5\,\text{m},5\,\text{m})$.
Only $x = 10\,\text{m}$ matters:
\[ E_x = \frac{40}{10^2} = \frac{40}{100} = 0.40\,\text{V\,m}^{-1} \]

Step 5: Write the complete electric field vector.
\[ \vec{E} = 0.40\,\hat{i}\,\text{V\,m}^{-1} \] The field points in the positive $x$-direction because the potential decreases as $x$ increases (higher $x$, lower $\phi$).

Step 6: State the answer.
\[ \boxed{0.40\,\text{V\,m}^{-1}\,\hat{i}} \]