# Solution to the Question Let's clarify and break down your question step by step: You are given a **cube-shaped sample** with certain dimensions and electric properties. You are to find: - The **magnitude of current** given the drift velocity. - The **electric field intensity**. - The **potential at point P**. - The **electric field at point P**. - The **power at point P**. - The **equation of the equipotential surface passing through point P**. I'll answer each sub-question, with example values and steps. A graph will be included at the end. --- ## Step 1: Understand the Given Data Let’s extract the data from your question: - **Sample shape:** Cube - **Side length of cube (a):** - 3.5 mm (for part c) - 3.6 mm (for part d) - **Potential difference (V):** .304 V to .5 mV (exact value unclear, will use .304 V as example) - **Drift velocity ($v_d$):** 190 cm/s ($= 19$ m/s) - **Electric field intensity ($E$):** 500 μV/m = $5 \times 10^{-4}$ V/m - **Total current ($I$):** 800 mA = .8 A (for part d) --- ## Step 2: Solve Each Part ### (a) Magnitude of the Current, Given Drift Velocity **Formula:** \[ I = n e v_d A \] - $n$: number density of charge carriers - $e$: elementary charge ($1.6 \times 10^{-19}$ C) - $v_d$: drift velocity - $A$: cross-sectional area **Assuming** $n = 8.5 \times 10^{28} \mathrm{~m}^{-3}$ (typical for copper) Cube side $a = 3.5\,\text{mm} = 3.5 \times 10^{-3}\,\text{m}$ Area $A = a^2 = (3.5 \times 10^{-3})^2 = 1.225 \times 10^{-5}\,\text{m}^2$ Drift velocity $v_d = 19\,\text{m/s}$ **Calculation:** \[ I = n e v_d A = (8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times 19 \times (1.225 \times 10^{-5}) \] \[ I \approx (8.5 \times 1.6 \times 19 \times 1.225) \times 10^{28-19-5} \] \[ I \approx (8.5 \times 1.6 = 13.6; 13.6 \times 19 = 258.4; 258.4 \times 1.225 \approx 316.54) \] \[ I \approx 316.54 \times 10^{28-19-5} = 316.54 \times 10^{4} \] \[ I \approx 3.165 \times 10^{6}\,\text{A} \] **Note:** This value is unphysically large for a small sample; the drift velocity or $n$ value may be off for the example. For principle, the _formula and steps_ are shown. --- **Example with smaller $n$ (e.g., $n = 8.5 \times 10^{22}$):** \[ I = (8.5 \times 10^{22}) \times (1.6 \times 10^{-19}) \times 19 \times (1.225 \times 10^{-5}) \] \[ = 3.165 \times 10^{} = 3.17\,\text{A} \] --- ### (b) Electric Field Intensity Given $E = 500\,\mu\text{V/m} = 5 \times 10^{-4}\,\text{V/m}$ --- ### (c) Potential at Point P If one face is at $V_1 = .304$ V and the opposite at $V_2 = $ V, and $P$ is $x$ mm from the $V_2$ face: \[ V(x) = V_2 + \frac{x}{a}(V_1 - V_2) \] - At $x=$: $V=\,\text{V}$ - At $x=a$: $V=.304\,\text{V}$ - At $x = a/2 = 1.75\,\text{mm}$: \[ V(a/2) = + \frac{1.75}{3.5}(.304) = .152\,\text{V} \] --- ### (d) Equation of Equipotential Surface For a cube aligned along $x$, equipotential surfaces are planes perpendicular to $x$: \[ V(x) = \text{constant} \] So, \[ x = k, \quad \text{where } k \text{ is a constant for a given potential} \] --- ### (e) Electric Field at P The electric field in a uniform cube: \[ E = -\frac{dV}{dx} = \frac{\Delta V}{\Delta x} \] \[ E = \frac{.304\,\text{V}}{3.5 \times 10^{-3}\,\text{m}} = 86.9\,\text{V/m} \] --- ### (f) Power at P Power at P (total power delivered): \[ P = I \cdot V \] \[ P = .8\,\text{A} \times .304\,\text{V} = .243\,\text{W} \] --- ## Example Summary Table | Quantity | Formula | Value (Example) | |-----------------|----------------------------------|--------------------------| | Current ($I$) | $n e v_d A$ | $3.17$ A (see above) | | Electric Field ($E$) | $\Delta V/\Delta x$ | $86.9$ V/m | | Potential at $a/2$ | $V(x) = \frac{x}{a} V_1$ | $.152$ V | | Power ($P$) | $I \cdot V$ | $.243$ W | | Equipotential | $x = \text{constant}$ | Plane at $x = k$ | --- ## Graph: Potential vs. Distance ```mermaid graph LR A[ mm, V] --|linear| B[1.75 mm, .152 V] --|linear| C[3.5 mm, .304 V] ``` **Alt text:** Linear graph of potential vs. distance across the cube, starting at V at mm and rising to .304 V at 3.5 mm. --- ## Final Answers (Examples) 1. **Current:** $I \approx 3.17$ A (with $n = 8.5 \times 10^{22} \mathrm{~m}^{-3}$) 2. **Electric field:** $E = 86.9$ V/m 3. **Potential at $x = a/2$:** $V = .152$ V 4. **Power:** $P = .243$ W 5. **Equipotential surface equation:** $x = \text{constant}$ --- **Note:** Adjust calculations if you have the actual value of $n$ or more precise values. Steps above show the general method.