Electric Potential: Practice Problem & Solution
If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$, the electric field (in N/C) at point $(1, 1, 0)$ is: (2015)
Solution Explained:
To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:
Electric field $E = -\nabla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k})$. $\frac{\partial V}{\partial x} = 6y$, $\frac{\partial V}{\partial y} = 6x - 1 + 2z$, $\frac{\partial V}{\partial z} = 2y$. At point $(1, 1, 0)$, $E_x = -6(1) = -6$, $E_y = -(6(1) - 1 + 0) = -5$, $E_z = -2(1) = -2$. Thus, $E = -(6\hat{i} + 5\hat{j} + 2\hat{k})$.
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