Question

In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

• A. \(10\ \text{N/C}\) along \(+x\)
• B. \(10\ \text{N/C}\) along \(-x\)
• C. \(50\ \text{N/C}\) along \(+x\)
• D. \(50\ \text{N/C}\) along \(-x\)

Hint:

Whenever the potential is given as a function of position, use \[ E=-\frac{dV}{dx}. \] A positive value of \(E\) means the field is along \(+x\), whereas a negative value means it is along \(-x\).

Answer 1

The Correct Option is C

Concept: The electric field and electric potential are closely related quantities in electrostatics. The electric field at any point is defined as the negative gradient of the electric potential. For one-dimensional motion along the \(x\)-axis, \[ E_x=-\frac{dV}{dx} \] The negative sign indicates that the electric field always points in the direction of decreasing potential. Therefore, to determine the magnitude and direction of the electric field, we simply differentiate the given potential function with respect to \(x\).

Step 1: Write the given expression for electric potential.
The potential is given as \[ V=10-50x \] where \(V\) is measured in volts and \(x\) in meters.

Step 2: Differentiate the potential with respect to \(x\).
\[ \frac{dV}{dx}=\frac{d}{dx}(10-50x) \] Since the derivative of a constant is zero, \[ \frac{dV}{dx}=0-50=-50 \] Thus, \[ \frac{dV}{dx}=-50\ \text{V/m} \]

Step 3: Use the relation between electric field and potential gradient.
\[ E=-\frac{dV}{dx} \] Substituting the value, \[ E=-(-50)=50\ \text{V/m} \] Since \[ 1\ \text{V/m}=1\ \text{N/C}, \] we get \[ E=50\ \text{N/C} \] The positive sign indicates that the electric field is directed along the positive \(x\)-axis. \[ \boxed{E=50\ \text{N/C along } +x} \]