Question

Two identical parallel plates are given charges as shown in figure. If plate area is A & plate seperation is d then heat liberated after closing switch s is.

• A. $\frac{5Q^{2}d}{{\epsilon }_{0}A}$
• B. $\frac{9Q^{2}d}{2{\epsilon }_{0}A}$
• C. $\frac{7Q^{2}d}{3{\epsilon }_{0}A}$
• D. $\frac{3Q^{2}d}{2{\epsilon }_{0}A}$

Answer 1

The correct option is B $\frac{9Q^{2}d}{2{\epsilon }_{0}A}$

The charge distribution will be as shown in figure.
Hence, heat generated $=\frac{(3Q{)}^{2}d}{2{\epsilon }_{0}A}$
Heat = $\frac{9Q^{2}d}{2{\epsilon }_{0}A}$