Parallel Plates and Electric Field

  • #1

Homework Statement



The electrostatic potential energy of a system of point charges q1 = 1 µC, q2 = 2 µC, and q3 = 3 µC at the corners of an equilateral triangle whose side s = 30 cm is


a. 1.10 J

b. 0.990 J

c. 0.631 J

d. 0.330 J

e. 0.123 J



Homework Equations



See below.

The Attempt at a Solution



PE = k*q*q_0/(r)

PE_total = [(k*(1*10^-6 C)*(3*10^-6 C)/(.30 m)] + [(k*(1*10^-6 C)*(2*10^-6 C)/(.30 m)] + [(k*(3*10^-6 C)*(2*10^-6 C)/(.30 m)]

= k/.30 m*[(3*10^-12 C^2) + (2*10^-12) + (6*10^-12)] = 0.3296 J ??










Homework Statement




Two flat parallel plates are a distance d = 0.40 cm apart. The potential difference between the plates is 360 V. The electric field at a point midway between the plates is approximately


a. 90 kN/C

b. 3.6 kN/C

c. 0.9 kN/C

d. zero

e. 3.6*10^5 N/C

Homework Equations



See below.

The Attempt at a Solution



E = delta V/d = 360 V/0.004 m = 90,000 V/m = 90,000 N/C ?

Thanks.
 

Answers and Replies

  • #2
70
0
Your solutions appear correct to me.

Potential energy is the sum of the energys required to add each charge in sequence (when dealing with discrete charges). You have followed this rule in your calculations adding first the 1 microC, then the 3 microC then the 2 microC (this is implicit in the way you wrote out the equations). The order is not important since E fields superpose linearly (ie they add up in the simplest way).
The second problem also seems right to me. This is a uniform field problem and the calculated value of field strength is constant thoughout the region.
 

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