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: Write the relation between electric field and potential.
Electric field is related to electric potential by \[ \vec{E}=-\nabla \phi \] Given, \[ \phi(x,y,z)=\phi_0\frac{x_0}{x} \] Substituting \[ \phi_0=8\,V \] and \[ x_0=5\,m, \] we get \[ \phi=\frac{8\times 5}{x} \] \[ \phi=\frac{40}{x} \]

Step 2: Differentiate the potential with respect to \(x\).
Since the potential depends only on \(x\), \[ E_x=-\frac{d\phi}{dx} \] Now, \[ \frac{d\phi}{dx}=\frac{d}{dx}\left(\frac{40}{x}\right) \] \[ \frac{d\phi}{dx}=-\frac{40}{x^2} \] Therefore, \[ E_x=-\left(-\frac{40}{x^2}\right) \] \[ E_x=\frac{40}{x^2} \] Hence, \[ \vec{E}=\frac{40}{x^2}\hat{i} \]

Step 3: Evaluate the electric field at \(x=10\,m\).
At the point \[ (10\,m,5\,m,5\,m), \] we have \[ x=10\,m \] So, \[ \vec{E}=\frac{40}{(10)^2}\hat{i} \] \[ \vec{E}=\frac{40}{100}\hat{i} \] \[ \vec{E}=0.40\,Vm^{-1}\hat{i} \]

Step 4: Final conclusion.
Hence, the electric field at the given point is \[ \boxed{0.40\,Vm^{-1}\,\hat{i}} \]