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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

A uniform field of 2.0 N C^ -1 exists in space in the x -direction....

A uniform field of 2.0 N C−12.0 \text{ N C}^{-1} exists in space in the xx-direction. Assuming the potential at the origin is zero, determine the expression for the potential at an arbitrary point (x,y,z)(x, y, z).

Options

  1. A−(2.0 V m−1)x-(2.0 \text{ V m}^{-1}) x
  2. B(2.0 V m−1)z(2.0 \text{ V m}^{-1}) z
  3. C−(2.0 V m−1)y-(2.0 \text{ V m}^{-1}) y
  4. D(2.0 V m−1)x(2.0 \text{ V m}^{-1}) x

Correct answer

A. −(2.0 V m−1)x-(2.0 \text{ V m}^{-1}) x

Step-by-step solution

The relation between electric potential and electric field is given by dV=−E⃗⋅dr⃗dV = -\vec{E} \cdot d\vec{r}

Given E⃗=2.0i^ N C−1\vec{E} = 2.0 \hat{i} \text{ N C}^{-1} and dr⃗=dxi^+dyj^+dzk^d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}

dV=−(2.0i^)⋅(dxi^+dyj^+dzk^)=−2.0dxdV = -(2.0 \hat{i}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k}) = -2.0 dx

Integrating from the origin (0,0,0)(0,0,0) to an arbitrary point (x,y,z)(x,y,z):

∫0VdV=−∫0x2.0dx\int_{0}^{V} dV = -\int_{0}^{x} 2.0 dx

V−0=−2.0xV - 0 = -2.0 x

Since 1 N C−1=1 V m−11 \text{ N C}^{-1} = 1 \text{ V m}^{-1}, the expression for the potential at (x,y,z)(x,y,z) is −(2.0 V m−1)x-(2.0 \text{ V m}^{-1}) x
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