Parallel plate capacitor and point charge problems

[n3ro]

1. parallel plate with capacitor of 20uF, plate is 0.5mm apart.

Find
1. Area of each plate
2. potential difference if there's a change of magnitude of 30uC on each plate.
3. stored energy
4. electric field between the two plates
5. charge density on the plate.


|_| |_|
|_____|
.5mm

I know $$C=\frac{Eo*A}{d}$$

so would $$A=\frac{Eo*C}{D}$$ or A=Q / Eo *E

A= (8.85*10^ -12 ) * 20 / (.5 * 10^ -3) = 3.54 * 10 ^7 ?


2. three point charges on the x-axis q1= (10uc) at x= -4m, q2= (-5uC) at the origin, q3= (-20uC) at x= 4m.


Q1(-4,0)... Q2(0,0)... Q3(4,0)
(+)----------(-)-----------(-)
10uC... -5uC... -20UC


Find
1. electric field at origin
2. force on q2


would electric field on orgin = Zero & force on q2 equal

$$F1= Ke\frac{q1|q2|}{r^2}$$ , $$F2=Ke\frac{|q3|q2|}{r^2}$$

$$F1=\frac{(8.99* 10^9)* (10 * 10^ -6)*( 5*10^-6)}{(-4)^ 2}$$

$$F2=\frac{(8.99* 10^9)* (20 * 10^ -6)*(5*10^-6)}{(4)^ 2}$$

After this would i add components ( x i + y j )

 

[dextercioby]

For the first problem,the area u computed is incorrect.You began with the correct formula,but somehow screwed up the arithmetics.What about the other 4 points...?

Daniel.

 

[thepatientmental]

or use Pythagorean theorem to find the magnitude of the force?

1. The area of each plate would be calculated as A = (8.85*10^-12 * 20*10^-6)/0.5*10^-3 = 3.54*10^-3 m^2.

2. To find the potential difference, we use the equation V = Q/C, where Q is the charge on each plate and C is the capacitance. Since the charge on each plate is 30 uC, the potential difference would be V = (30*10^-6)/20*10^-6 = 1.5 V.

3. The stored energy in a capacitor is given by E = 1/2 * C * V^2. Plugging in the values, we get E = 1/2 * 20*10^-6 * (1.5)^2 = 0.0225 J.

4. The electric field between two parallel plates is given by E = V/d, where V is the potential difference and d is the distance between the plates. Plugging in the values, we get E = (1.5)/(0.5*10^-3) = 3000 V/m.

5. The charge density on the plates is given by σ = Q/A, where Q is the charge on each plate and A is the area of the plates. Plugging in the values, we get σ = (30*10^-6)/(3.54*10^-3) = 8476.68 C/m^2.

For the three point charges on the x-axis, the electric field at the origin would be zero since the contributions from q1 and q3 cancel each other out.

To find the force on q2, we use the equation F = k * (q1*q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the two charges. Plugging in the values, we get F = (8.99*10^9) * (10*10^-6 * 5*10^-6)/4^2 = 11.24 N. The components of this force can be found using vector addition or the Pythagorean theorem.