In an isolated parallel-plate capacitor of capacitance C, the four surface have charges $Q_{1}, Q_{2}, Q_{3}$ and $Q_{4}$ as shown. The potential difference between the plates is :

$\frac{Q_{1} + Q_{2}}{C}$
$∣ ∣ ∣ \frac{Q_{2}}{C} ∣ ∣ ∣$
$∣ ∣ ∣ \frac{Q_{3}}{C} ∣ ∣ ∣$
$\frac{1}{C} [(Q_{1} + Q_{2}) - (Q_{3} - Q_{4})]$

Question

In an isolated parallel-plate capacitor of capacitance C, the four surface have charges $Q_{1}, Q_{2}, Q_{3}$ and $Q_{4}$ as shown. The potential difference between the plates is :

$\frac{Q_{1} + Q_{2}}{C}$
$∣ ∣ ∣ \frac{Q_{2}}{C} ∣ ∣ ∣$
$∣ ∣ ∣ \frac{Q_{3}}{C} ∣ ∣ ∣$
$\frac{1}{C} [(Q_{1} + Q_{2}) - (Q_{3} - Q_{4})]$

Toppr Admin · Accepted Answer

The potential difference between-xA0-the plates is depend on the charge on the inner face of the plates-xA0-For parallel plate capacitor- Q2-x2212-Q3 or Q3-x2212-Q2Thus the potential difference between the plates is V-Q2C- or V-Q3C-xA0-

In an isolated parallel-plate capacitor of capacitance C, the four surface have charges Q1,Q2,Q3 and Q4 as shown. The potential difference between the plates is :Q1+Q2C∣∣∣Q2C∣∣∣∣∣∣Q3C∣∣∣1C[(Q1+Q2)−(Q3−Q4)]

The potential difference between the plates is depend on the charge on the inner face of the plates. For parallel plate capacitor, Q2=−Q3 or Q3=−Q2Thus the potential difference between the plates is V=|Q2C| or V=|Q3C|

The potential difference between the plates is depend on the charge on the inner face of the plates.
For parallel plate capacitor, $Q_{2} = - Q_{3}$ or $Q_{3} = - Q_{2}$
Thus the potential difference between the plates is $V = | \frac{Q_{2}}{C} |$ or $V = | \frac{Q_{3}}{C} |$