Question

The relation between electric field \( E \) and electric potential \( V \) is:

• A. \( E = -\frac{dV}{dr} \)
• B. \( E = \frac{dr}{dV} \)
• C. \( E = \frac{dV^2}{dr} \)
• D. None of these

Hint:

Electric field is the rate of change of potential with distance: \[ E = -\nabla V \] In 1D: \( E = -\frac{dV}{dx} \)

Answer 1

The Correct Option is A

The electric field is related to the negative gradient of electric potential. In one-dimensional form, this relationship is expressed as: \[ E = -\frac{dV}{dr} \] This equation indicates that the electric field \( E \) is equal to the negative rate of change of electric potential \( V \) with respect to distance \( r \). The negative sign shows that the electric field points in the direction of decreasing electric potential. In other words, a positive test charge placed in the field will naturally move from a region of higher potential to a region of lower potential.