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Step-by-Step Solution
Step 1: Identify the Given Parameters
The problem states:
ā¢ Area of the parallel plate capacitor, $A = 200 \text{ cm}^2 = 200 \times 10^{-4} \text{ m}^2$
ā¢ Separation between the plates, $d = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}$
ā¢ Force of attraction between the plates, $F = 25 \times 10^{-6}\ \text{N}$
ā¢ Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \ \text{C}^2 \text{/(NĀ·m}^2)$
Step 2: Recall the Relevant Formula for the Force Between Plates
For a parallel plate capacitor, the electric field $E$ between the plates is given by
$$
E = \frac{Q}{A \,\epsilon_0},
$$
assuming a plate separation where edge effects can be neglected.
However, it is also known that the force on one plate is
$$
F = \frac{Q^2}{2\,A\,\epsilon_0},
$$
when the plates are isolated and each carries charge $Q$.
Step 3: Express the Charge $Q$ in Terms of $V$
The capacitance $C$ of a parallel plate capacitor is given by
$$
C = \frac{\epsilon_0 A}{d}.
$$
Therefore, the charge $Q$ on the plates when a potential difference $V$ is applied is
$$
Q = C V = \frac{\epsilon_0 A}{d} \, V.
$$
Step 4: Substitute $Q$ into the Force Expression
From the force expression $F = \frac{Q^2}{2\,A\,\epsilon_0}$, substitute $Q = \frac{\epsilon_0 A}{d} \, V$:
$$
F = \frac{1}{2\,A\,\epsilon_0}
\left(\frac{\epsilon_0 A}{d}\,V\right)^2
= \frac{\epsilon_0^2 A^2 V^2}{2\,A\,\epsilon_0\,d^2}
= \frac{\epsilon_0 A\,V^2}{2\,d^2}.
$$
Rearranging to solve for $V$ gives:
$$
V^2 = \frac{2 F \, d^2}{\epsilon_0 A}
\quad \Longrightarrow \quad
V = d \sqrt{\frac{2F}{\epsilon_0 A}}.
$$
Step 5: Substitute the Numerical Values
Substitute $F = 25 \times 10^{-6}\ \text{N}$, $d = 1.5 \times 10^{-2}\ \text{m}$,
$\epsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{(NĀ·m}^2)$, and $A = 200 \times 10^{-4}\ \text{m}^2$ into
$$
V = d \sqrt{\frac{2F}{\epsilon_0 A}}.
$$
Hence,
$$
V = 1.5 \times 10^{-2}
\sqrt{\frac{2 \times 25 \times 10^{-6}}{8.85 \times 10^{-12} \times 200 \times 10^{-4}}}.
$$
Step 6: Simplify the Expression
Carefully simplifying the expression yields approximately $250 \text{ V}$.
Numerically, one finds that the expression under the square root adjusts to a value that leads to about $250 \text{ V}$ as the final answer.
Final Answer
The emf $V$ is approximately $250 \text{ V}$.
