Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite each other.Between the electrodes but not near their edges,the electric field strength is 2.5 × 10⁶ N/C.What is the magnitude of the charge on each electrode? (ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)35 nC B)18 nC C)16 nC D)30 nC

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Quizplus · Accepted Answer

The magnitude of the charge on each electrode can be found using the formula for the electric field between two parallel plates: $E = \frac{\sigma}{\epsilon_0}$, where $E$ is the electric field strength, $\sigma$ is the surface charge density ($\frac{Q}{A}$), and $\epsilon_0$ is the permittivity of free space. Rearranging for $Q$, we get $Q = E \cdot A \cdot \epsilon_0$. Substituting the given values, $E = 2.5 \times 10^6 \, \text{N/C}$, $A = 4.0 \, \text{cm} \times 4.0 \, \text{cm} = 16 \, \text{cm}^2 = 16 \times 10^{-4} \, \text{m}^2$, and $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2$, we find $Q = 2.5 \times 10^6 \times 16 \times 10^{-4} \times 8.85 \times 10^{-12} = 35.4 \times 10^{-9} \, \text{C} = 35 \, \text{nC}$.

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