Electric Potential Energy: Practice Problem & Solution
A particle of mass \(m\) carrying charge \(q\) is initially kept at rest at the origin. A uniform electric field \(E\) along \(x\)-axis is switched on. What will be its kinetic energy when its coordinates are \((a, b)\)?
Solution Explained:
To solve this problem, we apply the core principles of Electric Potential Energy. Understanding the underlying formula is key to arriving at the correct answer below:
Work done by the electric field \(\vec{E} = E\hat{i}\) is given by \(W = q\vec{E} \cdot \vec{d} = qE\hat{i} \cdot (a\hat{i} + b\hat{j}) = qEa\). By work-energy theorem, \(K_f - K_i = W \implies K_f = qEa\).
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