Question

In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz, where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2$ coulomb situated at point $(1, 1, 1)$ is

• A. $ 6 \sqrt 5 \,N $
• B. 30 N
• C. 24 N
• D. $ 4 \sqrt 35 \,N $

Answer 1

The Correct Option is D

 To find the electric force experienced by a charge of \(2\) coulombs situated at the point \((1, 1, 1)\) in a potential field given by \(V(x, y, z) = 6x - 8xy - 8y + 6yz\), we can follow these steps:

1. First, understand that the electric force on a charge in an electric field \(\mathbf{E}\) is given by \(\mathbf{F} = q \mathbf{E}\), where \(q\) is the charge.
2. The electric field \(\mathbf{E}\) is the negative gradient of the potential \(V\), i.e., \(\mathbf{E} = -\nabla V\).
3. Calculate the gradient \(\nabla V\), which is given by partial derivatives of \(V\) with respect to \(x\), \(y\), and \(z\):
   • \(\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6x - 8xy - 8y + 6yz) = 6 - 8y\)
   • \(\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6x - 8xy - 8y + 6yz) = -8x - 8 + 6z\)
   • \(\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6x - 8xy - 8y + 6yz) = 6y\)
4. Thus, the electric field \(\mathbf{E}\) is:
   • \(\mathbf{E} = -\left( \left(6 - 8y\right) \mathbf{i} + \left(-8x - 8 + 6z\right) \mathbf{j} + \left(6y\right) \mathbf{k} \right)\)
5. Substitute \(x = 1\), \(y = 1\), \(z = 1\) into the equation for \(\mathbf{E}\):
   • \(\mathbf{E}(1, 1, 1) = -\left( (6 - 8 \times 1) \mathbf{i} + (-8 \times 1 - 8 + 6 \times 1) \mathbf{j} + (6 \times 1) \mathbf{k} \right)\)
   • \(\mathbf{E}(1, 1, 1) = -\left( (-2) \mathbf{i} + (-10) \mathbf{j} + 6 \mathbf{k} \right)\)
   • \(\mathbf{E}(1, 1, 1) = 2 \mathbf{i} + 10 \mathbf{j} - 6 \mathbf{k} \)
6. The force \(\mathbf{F}\) on the charge \(q = 2\) coulombs is:
   • \(\mathbf{F} = 2 \times (2 \mathbf{i} + 10 \mathbf{j} - 6 \mathbf{k}) = 4 \mathbf{i} + 20 \mathbf{j} - 12 \mathbf{k} \)
7. Calculate the magnitude of \(\mathbf{F}\) to find the electric force:
   • \(|\mathbf{F}| = \sqrt{(4)^2 + (20)^2 + (-12)^2}\)
   • \(|\mathbf{F}| = \sqrt{16 + 400 + 144}\)
   • \(|\mathbf{F}| = \sqrt{560}\)
   • \(|\mathbf{F}| = 4 \sqrt{35}\) N

Thus, the correct answer is \(4 \sqrt 35 \,N\).