- #1

- 160

- 0

## Homework Statement

two parallel plates have equal and opposite charges of 0.01 coulombs. the electric field between the plates is 10^5 newtons/coulomb and the separation between the plates is 2mm. what is the capacitance of the plates?

## Homework Equations

capacitance C = Q/V where Q is charge in coulombs, V is electric potential in volts

electric field E = V/d where d is distance in meters

Q = [epsilon_0*A/d]V where A is area and epsilon_0 is a constant = 8.85*10^-12

## The Attempt at a Solution

i solved the electric field equation for V, so that V = dE and substituted that dE for V in the capacitance equation to get:

C = Q/dE = Q/(0.002*10^5) = Q/200

my questions are how do i factor in two, opposite charges or two equal charges for that matter, into the equation or am i interpreting the problem incorrectly?

is the problem asking me to find the capacitance in two different cases?

case1 --> two plates, with the same charge

case2---> two plates, opposite charges

how can i use the equation C=Q/200 and incorporate the charges of TWO plates? there is only one Q in the equation.

if that is the case, will what i attempted earlier work to find the capacitance, it seems it will?

thanks

Last edited: