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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

An electric field E = ( i · 20 + j · 30) N C^ -1 exists in a region...

An electric field E⃗=(i⃗⋅20+j⃗⋅30) N C−1\vec{E} = (\vec{i} \cdot 20 + \vec{j} \cdot 30) \text{ N C}^{-1} exists in a region of space. Assuming the potential at the origin is taken to be zero, determine the potential at the point (2 m,2 m)(2 \text{ m}, 2 \text{ m}).

Options

  1. A−20 V-20 \text{ V}
  2. B−100 V-100 \text{ V}
  3. C−50 V-50 \text{ V}
  4. D100 V100 \text{ V}

Correct answer

B. −100 V-100 \text{ V}

Step-by-step solution

The relationship between electric potential and a uniform electric field is given by:

V−V0=−E⃗⋅r⃗V - V_0 = -\vec{E} \cdot \vec{r}

Given the electric field E⃗=20i^+30j^ N C−1\vec{E} = 20\hat{i} + 30\hat{j} \text{ N C}^{-1} and the position vector of the point r⃗=2i^+2j^ m\vec{r} = 2\hat{i} + 2\hat{j} \text{ m}.

Since the potential at the origin is V0=0V_0 = 0, the potential at the point (2,2)(2, 2) is:

V=−(20i^+30j^)⋅(2i^+2j^)V = -(20\hat{i} + 30\hat{j}) \cdot (2\hat{i} + 2\hat{j})

V=−(20×2+30×2)V = -(20 \times 2 + 30 \times 2)

V=−(40+60)=−100 VV = -(40 + 60) = -100 \text{ V}
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