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Electrostatic Sensor Array Potential Gradients (Electrical & Computer Engineering) Solution

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Problem Overview

We evaluate the 3D potential field V(x,y,z)=50+4x2−3y2+z2V(x, y, z) = 50 + 4x^2 - 3y^2 + z^2 at sensor point P(1,2,−2)P(1, 2, -2). We compute the potential gradient ∇V\nabla V, find the maximum rate of voltage increase ∥∇V∥\|\nabla V\|, calculate the electric field vector E⃗=−∇V\vec{E} = -\nabla V, and evaluate electrostatic force trajectories for MEMS nanoparticle sorting.

Step-by-Step Mathematical Solution

1. Compute Spatial Partial Derivatives and Voltage Gradient

State the 3D Gradient Vector Formula (defined in The Gradient Vector):

∇V(x,y,z)=⟨∂V∂x,∂V∂y,∂V∂z⟩=∂V∂xi^+∂V∂yj^+∂V∂zk^\nabla V(x, y, z) = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right\rangle = \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}

Compute the partial derivatives treating alternate variables as constants:

Vx=∂∂x[50+4x2−3y2+z2]=8xV_x = \frac{\partial}{\partial x}[50 + 4x^2 - 3y^2 + z^2] = 8x
Vy=∂∂y[50+4x2−3y2+z2]=−6yV_y = \frac{\partial}{\partial y}[50 + 4x^2 - 3y^2 + z^2] = -6y
Vz=∂∂z[50+4x2−3y2+z2]=2zV_z = \frac{\partial}{\partial z}[50 + 4x^2 - 3y^2 + z^2] = 2z

Evaluate at probe coordinates P(1,2,−2)P(1, 2, -2):

Vx(1,2,−2)=8(1)=8 V/cmV_x(1, 2, -2) = 8(1) = 8\text{ V/cm}
Vy(1,2,−2)=−6(2)=−12 V/cmV_y(1, 2, -2) = -6(2) = -12\text{ V/cm}
Vz(1,2,−2)=2(−2)=−4 V/cmV_z(1, 2, -2) = 2(-2) = -4\text{ V/cm}

Assemble into vector formats:

  • Bracket Notation: ∇V(1,2,−2)=⟨8,−12,−4⟩ V/cm\nabla V(1, 2, -2) = \langle 8, -12, -4 \rangle\text{ V/cm}
  • Unit Vector Notation: ∇V(1,2,−2)=8i^−12j^−4k^ V/cm\nabla V(1, 2, -2) = 8\hat{i} - 12\hat{j} - 4\hat{k}\text{ V/cm}

2. Maximum Spatial Rate of Change and Unit Vector Direction

The maximum rate of voltage increase is the Euclidean magnitude of the gradient:

∥∇V(1,2,−2)∥=82+(−12)2+(−4)2=64+144+16=224=414≈14.97 V/cm\|\nabla V(1, 2, -2)\| = \sqrt{8^2 + (-12)^2 + (-4)^2} = \sqrt{64 + 144 + 16} = \sqrt{224} = 4\sqrt{14} \approx 14.97\text{ V/cm}

The direction of maximum voltage increase as a unit vector is:

u⃗max=∇V∥∇V∥=⟨8,−12,−4⟩414=⟨214,−314,−114⟩≈⟨0.535,−0.802,−0.267⟩\vec{u}_{\text{max}} = \frac{\nabla V}{\|\nabla V\|} = \frac{\langle 8, -12, -4 \rangle}{4\sqrt{14}} = \left\langle \frac{2}{\sqrt{14}}, -\frac{3}{\sqrt{14}}, -\frac{1}{\sqrt{14}} \right\rangle \approx \langle 0.535, -0.802, -0.267 \rangle

3. Electric Field Vector and Physical Particle Trajectory

From electrostatics, the electric field vector is the negative potential gradient:

E⃗=−∇V=−⟨8,−12,−4⟩=⟨−8,12,4⟩ V/cm\vec{E} = -\nabla V = -\langle 8, -12, -4 \rangle = \langle -8, 12, 4 \rangle\text{ V/cm}

For an aerosol particle with positive charge q>0q > 0:

F⃗=qE⃗=q⟨−8,12,4⟩\vec{F} = q\vec{E} = q\langle -8, 12, 4 \rangle

  • Physical Direction: The Coulomb force drives the particle in the direction of E⃗=⟨−8,12,4⟩\vec{E} = \langle -8, 12, 4 \rangle, which is the path of steepest electrostatic potential decrease (steepest descent).
  • Engineering Application: To capture positively charged contaminant particles before they pass the sensor, electrical engineers orient grounded or negatively charged collection plates along the vector heading ⟨−8,12,4⟩\langle -8, 12, 4 \rangle.