Q145 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
An electric field $E=3 x^2 i N C^{-1}$ exists in a certain region of space. The potential difference between the origin and at $x=4 m, V_0-V_4$ is
- A$-20 \mathrm{~V}$
- Correct answerB64 V
- C$-64 \mathrm{~V}$
- D$-40 \mathrm{~V}$
Solution
The relationship between electric field $\vec{E}$ and electric potential $V$ is given by $dV = -\vec{E} \cdot d\vec{r}$.
Given $\vec{E} = 3x^2 \hat{i}$, we have $dV = -3x^2 dx$.
To find the potential difference $V_0 - V_4$, we integrate from $x=4$ to $x=0$:
$V_0 - V_4 = \int_{4}^{0} -3x^2 dx$.
$V_0 - V_4 = -3 \int_{4}^{0} x^2 dx$.
$V_0 - V_4 = -3 \left[ \dfrac{x^3}{3} \right]_{4}^{0}$.
$V_0 - V_4 = -[x^3]_{4}^{0} = -(0^3 - 4^3) = -(-64) = 64 \text{ V}$.
Answer: 64 V
Given $\vec{E} = 3x^2 \hat{i}$, we have $dV = -3x^2 dx$.
To find the potential difference $V_0 - V_4$, we integrate from $x=4$ to $x=0$:
$V_0 - V_4 = \int_{4}^{0} -3x^2 dx$.
$V_0 - V_4 = -3 \int_{4}^{0} x^2 dx$.
$V_0 - V_4 = -3 \left[ \dfrac{x^3}{3} \right]_{4}^{0}$.
$V_0 - V_4 = -[x^3]_{4}^{0} = -(0^3 - 4^3) = -(-64) = 64 \text{ V}$.
Answer: 64 V
Hence, the correct answer is option: 'B'.