Question

Force of attraction between the plates of a parallel plate capacitor is

• A. $\frac { q ^ { 2 } } { 2 \varepsilon _ { 0 } A K }$
• B. $\frac { q ^ { 2 } } { \varepsilon _ { 0 } A K }$
• C. $\frac { q } { 2 \varepsilon _ { 0 } A }$
• D. $\frac { q ^ { 2 } } { 2 \varepsilon _ { 0 } A ^ { 2 } K }$

Answer 1

Answer: (A)

$\frac { q ^ { 2 } } { 2 \varepsilon _ { 0 } A K }$

Answer 2

Force on one plate due to another is

F = qE = $= q \left( \frac { q } { 2 A K \varepsilon _ { 0 } } \right) = \frac { q ^ { 2 } } { 2 A K \varepsilon _ { 0 } }$

(where $\frac { \sigma } { 2 \varepsilon _ { 0 } K }$ is the electric field produced by one plate at the location of other).