Chapter 23: Q. 12 (page 652)

A parallel-plate capacitor consists of two square plates, size $L \times L,$ separated by distance d. The plates are given charge $\pm Q$ . What is the ratio $E_{f} / E_{i}$ of the final to initial electric field strengths if
(a) Q is doubled,
(b)L is doubled, and
(c) d is doubled? Each part changes only one quantity; the other quantities have their initial values

Short Answer

Expert verified

(a) The ratio of the final to initial electric field strengths, if Q is doubled, is also doubled.

(b) The ratio of the final to initial electric field strengths if L is doubled is $\frac{1}{4}$ .

Step by step solution

Given information (part a)

$C h a r g e Q_{i} = Q A r e a A_{i} = L \times L C h a r g e Q_{i} = 2 Q A r e a A_{f} = L \times L$

Explanation (part a)

$E = \frac{charge}{4 π e_{0} \times Area} \Rightarrow E_{1} = \frac{Q_{i}}{4 π ε_{0} \times A_{i}} \Rightarrow E_{2} = \frac{Q_{f}}{4 π e_{0} \times A_{f}} \Rightarrow \frac{E_{2}}{E_{1}} = \frac{\frac{Q_{f}}{4 s_{e_{i}} \times A_{f}}}{\frac{Q_{i}}{4 π ε_{0} \times A_{i}}} Substitutingthegivendata \Rightarrow \frac{E_{2}}{E_{1}} = \frac{\frac{2 Q}{4 π e_{0} \times L \times L}}{\frac{Q}{4 π ε_{0} \times L \times L}} Sinceareaissameinboththecasesandhencetheycancelout \Rightarrow \frac{E_{2}}{E_{1}} = 2$

Given information (part b)

$C h a r g e Q_{i} = Q A r e a A_{i} = L \times L C h a r g e Q_{i} = Q A r e a A_{f} = 2 L \times 2 L$

Explanation (part b)

$E = \frac{charge}{4 π e_{0} \times Area} \Rightarrow E_{1} = \frac{Q_{i}}{4 π ε_{0} \times A_{i}} \Rightarrow E_{2} = \frac{Q_{f}}{4 π e_{0} \times A_{f}} \Rightarrow \frac{E_{2}}{E_{1}} = \frac{\frac{Q_{f}}{4 s_{e_{i}} \times A_{f}}}{\frac{Q_{i}}{4 π ε_{0} \times A_{i}}} \Rightarrow \frac{E_{2}}{E_{1}} = \frac{\frac{Q}{4 π ε_{0} \times 2 L \times 2 L}}{\frac{Q}{4 π ε_{0} \times L \times L}} \Rightarrow \frac{E_{2}}{E_{1}} = \frac{1}{4}$