THE PRACTICE STUDIO
Go slowly. Ask why. Make it yours.
PRACTICE / Electric field from potential
Differentiate potential with its sign
In a region, electrostatic potential depends only on x and is V(x) = 12 − 3x² volts when x is measured in metres.
Find Ex at x = 2 m.
Organize the givens & assumptions
- V(x) = 12 − 3x² V
- x = 2 m
Electrostatic differentiable potential; coefficients carry the SI units required by V.
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Problem: Differentiate potential with its sign
In a region, electrostatic potential depends only on x and is V(x) = 12 − 3x² volts when x is measured in metres.
Question: Find Ex at x = 2 m.
Givens:
- V(x) = 12 − 3x² V
- x = 2 m
Assumptions: Electrostatic differentiable potential; coefficients carry the SI units required by V.
Choices (LaTeX):
A. -12\ \mathrm{V/m}
B. 0\ \mathrm{V/m}
C. +6\ \mathrm{V/m}
D. +12\ \mathrm{V/m}
Ask for my approach first. Help me distinguish a physical misconception from an algebra error without revealing which choice is correct.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.Start with the smallest hint.
Work through it with me.
One decision at a time. Pause, predict, then unpack the algebra.
Translate a position-dependent potential into a local field.
Use the local field relation
A field points in the direction of decreasing potential.
Why is this step valid?
The reference value of potential does not determine the local force per charge.
Differentiate before evaluating
The constant 12 disappears while x² contributes 2x.
Why is this step valid?
The derivative supplies volts per metre and the minus sign sets the direction.
Evaluate the coordinate
The field points along positive x at this point.
Why is this step valid?
The positive result is consistent with potential decreasing as x increases near x = 2 m.
A symmetric finite difference has the expected negative potential slope.
TRY THE IDEA AGAIN
Adding 50 V everywhere changes the field here to…
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