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 a

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 meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is

Accepted Answer

1. **Relationship between Field and Potential:** The electric field $\mathbf{E}$ is the negative gradient of the potential $V$ [1].
   $\mathbf{E} = -
abla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} ight)$

2. **Partial Derivatives Calculation:**
   Given $V = 6x - 8xy - 8y + 6yz$
   $\frac{\partial V}{\partial x} = 6 - 8y$
   $\frac{\partial V}{\partial y} = -8x - 8 + 6z$
   $\frac{\partial V}{\partial z} = 6y$

3. **Electric Field at Point (1, 1, 1):**
   Substitute $x=1, y=1, z=1$:
   $E_x = -(6 - 8(1)) = -(-2) = 2$
   $E_y = -(-8(1) - 8 + 6(1)) = -(-10) = 10$
   $E_z = -(6(1)) = -6$
   $\mathbf{E} = 2\hat{i} + 10\hat{j} - 6\hat{k} 	ext{ V/m}$

4. **Electric Force Calculation:**
   The force experienced by a charge $q$ in an electric field is $\mathbf{F} = q\mathbf{E}$ [2].
   Given $q = 2 	ext{ C}$:
   $\mathbf{F} = 2(2\hat{i} + 10\hat{j} - 6\hat{k}) = 4\hat{i} + 20\hat{j} - 12\hat{k} 	ext{ N}$

5. **Magnitude of Force:**
   $|\mathbf{F}| = \sqrt{4^2 + 20^2 + (-12)^2} = \sqrt{16 + 400 + 144} = \sqrt{560}$
   $|\mathbf{F}| = \sqrt{16 	imes 35} = 4\sqrt{35} 	ext{ N}$

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