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Step-by-Step Solution
Step 1: Understand the Given Data
We have a parallel plate capacitor with capacitance
$1\,\mu\text{F}$. One plate is given a charge of
$+2\,\mu\text{C}$, while the other plate is given a charge of
$+4\,\mu\text{C}$. We wish to find the potential difference
across the capacitor.
Step 2: Recognize the Relevant Concept
A capacitor’s potential difference depends on the difference
in charge between its plates. However, when both plates have
charges of the same sign (both positive here), we decompose the
charge distribution into two parts:
An overall "average" charge that sits on both plates equally and does not create a potential difference.
An effective "excess" charge of opposite sign on each plate, which does create the potential difference.
If one plate carries $+q_1$ and the other $+q_2$, the potential
difference $V$ is related to the difference
$q_1 - q_2$. The formula for the potential difference becomes:
$$
V = \frac{|\,q_1 - q_2\,|}{2\,C}.
$$
Step 3: Apply the Formula to This Problem
Here:
$q_1 = 2\,\mu\text{C}$,
$q_2 = 4\,\mu\text{C}$,
and
$C = 1\,\mu\text{F}.$
Compute the difference in charges:
$$
|\,q_2 - q_1\,| = |\,4\,\mu\text{C} - 2\,\mu\text{C}\,| = 2\,\mu\text{C}.
$$
Therefore,
$$
V = \frac{2\,\mu\text{C}}{2 \times 1\,\mu\text{F}}
= \frac{2\,\mu\text{C}}{2\,\mu\text{F}}
= 1\,\text{V}.
$$
Step 4: State the Final Answer
The potential difference developed across the capacitor is
1 V.
