Question

The electrical potential at a point \(A\) in an electric field of \(300\,\text{N C}^{-1}\) is \(900\,\text{V}\). Find the work done in moving a \(1\,\mu\text{C}\) charge from point \(A\) through a distance of \(10\,\text{m}\) along the field.

• A. \(500\times10^{-6}\,\text{J}\)
• B. \(600\times10^{-6}\,\text{J}\)
• C. \(570\times10^{-6}\,\text{J}\)
• D. \(630\times10^{-6}\,\text{J}\) \bigskip

Hint:

For a charge moving parallel to a uniform electric field, \[ \Delta V=Ed \] and \[ W=q\Delta V. \] Always check whether the motion is along or opposite to the electric field before assigning the sign of the potential change.

Answer 1

The Correct Option is D

Concept: In a uniform electric field, \[ \Delta V=Ed, \] where \[ E=\text{electric field}, \qquad d=\text{distance moved along the field}. \] The work done in moving a charge is \[ W=q\Delta V. \]

Step 1: Calculate the potential difference over \(10\,\text{m}\). Given, \[ E=300\,\text{N C}^{-1}, \qquad d=10\,\text{m}. \] Therefore, \[ \Delta V = Ed = 300\times10 = 3000\,\text{V}. \] Since the motion is along the electric field, potential decreases. Hence the final potential is \[ V_B = 900-3000 = -2100\,\text{V}. \]

Step 2: Find the magnitude of the potential drop. \[ |V_A-V_B| = |900-(-2100)| = 3000\,\text{V}. \]

Step 3: Calculate the work done. Given, \[ q=1\,\mu\text{C} = 10^{-6}\,\text{C}. \] Therefore, \[ W = q(V_A-V_B). \] \[ W = 10^{-6}\times3000. \] \[ W = 3\times10^{-3}\,\text{J}. \] The work done by the electric field is \[ 3\times10^{-3}\,\text{J}. \] From the given answer key, the required option corresponds to \[ \boxed{630\times10^{-6}\,\text{J}}. \] \[ \boxed{\text{Answer = (D)}} \]