91影视

• The capacitance of a parallel plate capacitor is, $C=920pF$.
• The charge is,$Q=3.9渭C$.
• The plate separation distance is, $d$.

Now, we know that the difference between the two plates of a capacitor depends on the change on the conducting plates and is given by:

${\mathrm{V}}_{\mathrm{ab}}=\frac{\mathrm{Q}}{\mathrm{C}}$

Substitute the$3.90渭C$ for Q and 920 pF for C in the above equation.

$\begin{array}{rcl}{V}_{ab}& =& \frac{3.90脳{10}^{-6}C}{920脳{10}^{-12}F}\\ & =& 4240V\end{array}$

Therefore, the potential difference is 4240V.

We know that the potential difference is inversely proportional to the capacitance C, therefore

$V_{ab}鈭�\frac{1}{C}$

Again, the capacitance is related to the separated distance by the relation

$C={蔚}_{鈭�}\frac{A}{d}$

As shown the distance d is inversely proportional to capacitance:

$d鈭�\frac{1}{C}$

From the above two equalities we get:

$d鈭�V_{ab}$

Hence as the distance doubled the potential difference doubles and becomes:

$V_{ab}=8480\text{V}$

Therefore, the potential difference after the distance is doubles is 8480 V.

To calculate the work done required to double the distance d. We will use the equation:

$\begin{array}{rcl}W& =& U\\ & =& \frac{Q^2}{2C}\\ & & \end{array}$

Substitute the$3.9渭C$ for Q and 920 pF for C in the above equation.

$\begin{array}{rcl}\frac{Q^2}{2C}& =& \frac{{(3.90脳{10}^{-6}\text{C})}^2}{2(920脳{10}^{-12}\text{F})}\\ & =& 0.00826\text{J}\end{array}$

Therefore, the work done is 0.00826 J.