Problem 7: Consider a parallel-plate capacitor made up of two conducting plates
ID: 1792467 • Letter: P
Question
Problem 7: Consider a parallel-plate capacitor made up of two conducting plates with dimensions 23 mm x 15 mm. Part (a) If the separation between the plates is 1.6 mm, what is the capacitance, in pF, between them? Numeric : A numeric value is expected and not an expression Part (b) If there is 0.51 nC of charged stored on the positive plate, what is the potential, in volts, across the capacitor? Numeric : A numeric value is expected and not an expression - Part (c) What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? Numeric :A numeric value is expected and not an expression Part (d) If the separation between the plates doubles, what will the electric field be if the charge is kept constant? Numeric :A numeric value is expected and not an expression.Explanation / Answer
Though the questions require Numerical answers, I'll help you with expressions you can use to find these values
a) We can find the Capacitance by C = 0 A/d, 0 is permittivity of free space, A is area of plates and d is the dist between them. This comes to be 1.91 pF
b) Thevold difference is goven by V = QA/0 d where Q is the total charge and others have meanings same as above. If we calculate, we'll get voltage diff= 267.13 V
c) Electric field between the plates is E = Q/A0 where symbols have the same meanings as above. Upon simlple calculation, we get electric field as 1.67x105 N/C
d) No Change in elect field because as per the relationship in c) above, the Elect field doesnt depend on separation between plates. It only depends on the charge density (Q/A) on the parallel plates