Question

Assume that an electric field \(\vec{E}=30x^{2}\hat{i}\) N/C exists in space. If \(V_{0}\) is the potential at the origin and \(V_{A}\) is the potential at \(x=2\) m, then the potential difference \(V_{A}-V_{0}\) is

• A. \(-80\ V\)
• B. \(80\ V\)
• C. \(-120\ V\)
• D. \(120\ V\)

Hint:

Integrate the field along x with a negative sign to get the potential difference.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(dV=-\vec{E}\cdot d\vec{x}\), so \(V_{A}-V_{0}=-\int_{0}^{2}E\,dx\).

Step 2: Calculation
\(V_{A}-V_{0}=-\int_{0}^{2}30x^{2}\,dx=-30\left[\dfrac{x^{3}}{3}\right]_{0}^{2}=-10\times8=-80\ \text{V}\).

Step 3: Sign check
Potential falls in the direction of the field, so a negative value is expected.

Final Answer:
The potential difference is minus 80 volt. \[ \boxed{\text{(A) }-80\ V} \]