This is a link to the homework i am attempting to do.. http://www4.wittenberg.ed

Question

This is a link to the homework i am attempting to do.. http://www4.wittenberg.edu/academics/phys/pvoytas/p21806/labs/lab5_cap/caphw.pdf I understand the basics of capacitance but i cant find any help online or in my book about what happens once the two uncharged metal plates are connected and am not sure how this affects everything??? ie an expression for the work done by the electric field, magnitude and direction of the new electric field, and the charge density...im not sure how the plates change it if at all?? Please help with some explanations that will help me complete this

Explanation / Answer

1. a. C = e0 A/d

b. i) potential is set by battery so unchanged

ii) W = d PE = q dV = q V
iii) E= V/d
independent of A so unchanged
iv) Q = CV = e0 A/d V
Q/A = e0 V/d
sigma = e0 V/d

doesnt depend on A so unchanged

v) Q = e0 A V/d
this depends on A so increasing A increased Q

vi) C = e0 A/d
increasing A incrased C

2
i) Q/A = Q1/( A1 + A2)
ii) Q = CV
V = Q/C = Q/ ( e0 ( A1 + A2)/d)

iii) C = e0 (A1 + A2)/d
which is the same as part 1 since capacitance is a geometric and physics property

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.