- #1
n3ro
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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 [tex]C=\frac{Eo*A}{d}[/tex]
so would [tex]A=\frac{Eo*C}{D}[/tex] 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
[tex]F1= Ke\frac{q1|q2|}{r^2}[/tex] , [tex]F2=Ke\frac{|q3|q2|}{r^2}[/tex]
[tex]F1=\frac{(8.99* 10^9)* (10 * 10^ -6)*( 5*10^-6)}{(-4)^ 2}[/tex]
[tex]F2=\frac{(8.99* 10^9)* (20 * 10^ -6)*(5*10^-6)}{(4)^ 2}[/tex]
After this would i add components ( x i + y j )
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 [tex]C=\frac{Eo*A}{d}[/tex]
so would [tex]A=\frac{Eo*C}{D}[/tex] 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
[tex]F1= Ke\frac{q1|q2|}{r^2}[/tex] , [tex]F2=Ke\frac{|q3|q2|}{r^2}[/tex]
[tex]F1=\frac{(8.99* 10^9)* (10 * 10^ -6)*( 5*10^-6)}{(-4)^ 2}[/tex]
[tex]F2=\frac{(8.99* 10^9)* (20 * 10^ -6)*(5*10^-6)}{(4)^ 2}[/tex]
After this would i add components ( x i + y j )