In an isolated parallel plate capacitor of capacitance C the four surfaces have charges $Q_{1}, Q_{2}, Q_{3}$ and $Q_{4}$ as shown in the figure. 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 surfaces have charges $Q_{1}, Q_{2}, Q_{3}$ and $Q_{4}$ as shown in the figure. 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

-For parallel plate-capacitor- Q2 andQ3 must be same and equal in-magnitude-i-e-Q2-Q3-So- potential-difference between the plates- V-Q2-C-Q3-C-Ans- -B-C-

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

For parallel plate capacitor, Q2 andQ3 must be same and equal in magnitude.i.e,|Q2|=|Q3| So, potential difference between the plates, V=|Q2|C=|Q3|C Ans: (B),(C)

For parallel plate capacitor, $Q_{2}$ and $Q_{3}$ must be same and equal in magnitude.i.e, $| Q_{2} | = | Q_{3} |$

So, potential difference between the plates, $V = \frac{| Q_{2} |}{C} = \frac{| Q_{3} |}{C}$

Ans: $(B), (C)$