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NEET PHYSICSEasy

In a certain region of space with volume 0.2 m30.2 \text{ m}^3, the electric potential is found to be 5 V5 \text{ V} throughout. The magnitude of the electric field in this region is:

A

0.5 N/C0.5 \text{ N/C}

B

1 N/C1 \text{ N/C}

C

5 N/C5 \text{ N/C}

D

zero

Step-by-Step Solution

The electric field E\mathbf{E} and electric potential VV are related by the equation E=dVdr\mathbf{E} = -\frac{dV}{dr} (in one dimension) or E=V\mathbf{E} = -\nabla V (gradient of potential) [NCERT Class 12, Physics Part I, Sec 2.6, Eq 2.20].

  1. Given: The electric potential VV is 5 V5 \text{ V} (a constant value) throughout the volume.
  2. Concept: The electric field represents the rate of change of potential with respect to distance. Since the potential is constant, it does not change with position.
  3. Calculation: The derivative of a constant with respect to position is zero. E=d(5)dr=0E = -\frac{d(5)}{dr} = 0 Therefore, the electric field in a region of constant potential is always zero. This is analogous to the field inside a charged conductor where the potential is constant and the field is zero [NCERT Class 12, Physics Part I, Sec 2.9].
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